MidsummerSnowstorm - New York Universitygottlieb/tr/back-issues/1980s/1985/1985...MidsummerSnowstorm Ourgroup is just nowmovingto anewlyrenovatedfloor. Wehad a major influence in specifying

PUZZLE CORNER

ALLAN J. GOTTLIEB

Midsummer Snowstorm

Our group is just now moving to

a newly renovated floor. We had

a major influence in specifying

the layout, and it is satisfying to see how

well it all works. I guess that if our re

search efforts into computer architecture

turn out to fizzle we have a related field

to move into.

Problems

A/S 1. Here is a chess problem from The

Tech, M.I.T.'s student newspaper.

White is to move and force a mate in

two:

A/S 2. Another geometry problem from

Phelps Meaker, who writes:

I wanted an approximation of a four-foot

sphere for light-integration measure

ment. I made a dodecahedron of sheet

metal, except that instead of flat sur

faces, I substituted low five-sided pyr

amids. It served my purpose. How

should I have designed the pyramids so

that all the dihedral angles were equal?

A/S 3. Randy Barron poses a question

that should appeal to all EE majors:

SEND PROBLEMS. SOLUTIONS,

AND COMMENTS TO ALLAN /.

GOTTLIEB. '67, ASSOCIATE

RESEARCH PROFESSOR AT

THE COURANT INSTITUTE OF

MATHEMATICAL SCIENCES.

NEW YORK UNIVERSITY. 25]

MERCER ST.. NEW YORK.

NY.. 10012.

A four-port device contains only passive

linear circuit elements. Using a fixed-fre

quency sine-wave generator, you can

measure the complex impedance Z be

tween two of the terminals, A and B. A

variable resistor R is connected across

the other two terminals, C and D. What

is the locus in the complex plane traced

out by Z as R varies from zero (short

circuit) to infinity (open circuit)?

A/S 4. Here is a temporal cryptarithmetic

problem from Nob Yoshigahara relayed

to me via Richard Hess (on DICKNET in-

terpuzzler communication network):

Fill in the boxes with the digits 1,2,3.. .,9:

DD^OD" x □ = □□""": DD*"

A/S 5. An out-of-season problem from

Bruce Calder:

It began to snow on a certain morning,

and the snow continued to fall steadily

throughout the day. At noon, a snow-

plow started to clear a road at a constant

rate in terms of the volume of snow re

moved per hour. The snowplow cleared

two miles by 2 p.m. and one more mile

by 4 p.m. At what time had the snow

storm begun?

Speed Department

SD 1. Greg Huber sent us the following

problem, told to him by Douglas Hof-

stadter:

Simplify the product

(x - a)(x - b) . . . (x - z).

SD2. Doug Van Patter offers a bridge

quickie:

North: East:

AK93 A10 76

¥653 ¥ AKQ7

♦ AJ92 ♦ Q75

*Q74 *J109

You are East in a rubber bridge game;

bidding has gone: South INT, North

3NT. Your partner leads the ¥10, and

everyone follows to three rounds of

hearts. What is your best shot at setting

this game?

Solutions

APR 1. White is lo play and mate in two:

Jerry Grossman sent us the following solution:

White moves g5-f6 (en passant), then f6-f7 (check

mate) after any reply by Black. To prove that this

solution is correct, we first claim that Black's pre

vious move was either d7-d5 or (7-f5. These are

clearly the only pawn moves possible (since the

White king could not have been in check when Black

moved), but why couldn't the Black king have

moved to its present position on the last move to

escape check? It couldn't have come from d8 or f8,

since there would have been no way for White to

administer the double check (pawn at e7 and

knight). On the other hand, if the Black king came

from d7 or f7, Ihen White would have had to have

moved the pawn to e6 on the previous move but

there is no now-vacant square from which this

pawn could have come. Thus we have verified this

first claim. Now, in order for White's pawns to be

in the position they are in, they must have made at

least 10 captures. Black is missing only 10 pieces,

however, and hence every one must have been cap

tured by a pawn. If Black's last move was d7-d5,

then Black's queen bishop could not have ever

moved and could not have been captured by a

pawn. This contradiction proves that Black's last

move could only have been f7-f5, and thus the so-

AJO AUGUST/SEPTEMBER 1985

Clyde B. Dooiitlle, '23; Februar>- 22,1985; Hillsdale,N.J.

Thomas B. Drew, '23; May 5, 1985; Peterborouch,N.H. hWilliam J. O'Shaughnessey, '23; July 22, 1984; Ma-con, Ga.

Clinton B. Conway, '24; April 11, 1985; Altoona.Fla.

Robert Bruce Lindsay, '24; February 2, 1985; Providence, R.I.

Mrs. Sidney W. Andrews, '25; January 25, 1985;South Salem, N.Y.

Alexander C. Brown, '25; January 13. 1985; Cincinnati, Ohio.

George L. Bums, '25; 1984; Portland, Ore.

Frederic S. Jacques, '25; March 22, 1985; Bangor,Maine.

Augustine J. Cotter, '26; 1984; Scarsport, Maine.Valentine F. Harrington, '26; 1985.

Richard L. O'Donovan, '27; April 23, 1985; CoralGables, Ha.

Helmuth G.R. Schneider, '27; February 25, 1985-Westfield, N.J.

Darcy A. Young, Jr., "27; November 24, 1984; Peterborough, N.H.

Alexander B. Daytz, '28; December 1984; Chats-worth, Calif.

Mrs. David Mathoff, '28; 1985; Boston, Mass.Harold G. Nyman, '28; June 2,1984; Belmont, Mass.

Waller R. Ramsaur, '28; May 11, 1984; Pacific Palisades, Calif.

Joseph D. Riley, '28; March 19, 1985; LighthousePt., Fla.

Francis C. Sweeney, '28; March 15, 1985; Clifton.N.J.

Charles B. Bacon, '29; April 30, 1985; Middletown,Conn.

William E. Lowery, '29; November 29, 1984; Plymouth, Mass.

Henry S. Muller, '29; March 26, 1985; Belmont,Ohio.

Henry N. Bates, '30; January 1985; Los Gatos, Calif.

Thomas J. Hickey, '30; January 18, 1985; Arlington,Va.

Jonathan G. Swift, "30; January 18,1985; West Hartford, Conn.

Albert Earl Cullum, Jr., '31; January 31, 1985; Dallas, Tex.

Joseph H. Passell, '31; March 15,1985; Williamston,

N.CIra J. Bach, '32; March 6, 1985; Chicago, 111.

EdwinC. Beck, '32; May 31,1984; Muskegon, Mich.

Harry F. Carlson, '32; August 31, 1984; Hanover,Mass.

Maurice D. Triouleyre, '32; April 26, 1985; Long-meadow, Mass.

Joseph Welch, Jr., '32; April 13. 1985; Fort Lauder-dale, Fla.

Weston L. Brannen, '33; March 14,1978; Valparaiso,Ind.

J. Mason Culverwell, '33; November 3, 1984; Wash

ington, D.C.

Leo V. Dewar, '33; November 5. 1984; Rochester,

N.Y.

John R. Hopkins, '33; March 16, 1985; Los Alamos,N.M.

Kenneth D. Moslander, '33; February 1985; FortMyers, Fla.

Robert N. Eck, '34; March 29, 1985; Brookfield.Wise.

Waller R. Hedeman, Jr., '34; April 14, 1985; An

napolis, Md.

Octavio Leonard Colavecchio, '35; April 2, 1985;Providence, R.I.

Dwighl P. Merrill, '35; December 27,1984; Newton,

Mass.

Joseph 1. Ackerman, Jr., '36; May 9, 1985; Arling

ton, Mass.

Wilber W. Haynes, '38; 1985; Kissimmee, Fla.

Alfred Sweeney, Jr., '38; December 13, 1984; Santa

Barbara, Calif.

Robert G. Larkln, '39; February 15, 1985; VenturaCalif.

Ramon S. Sevilla, Sr., '39; April I, 1985; Metro Manila, Philippines.

Ernest G. Chilton, '40; January 22, 1985; MenloPark, Calif.

David Joseph Collins, '40; April 13,1985; Wcllesley,Mass.

J. Herbert Hollomon, '40; May 8, 1985; Brookline,Mass.

Spencer M. Richardson, '40; May 26. 1984; Quak-erlown, Penn.

Richmond W. Wilson, '40; February 7, 1985; Corning, N.Y.

Richard H. Gould, Jr., '41; 1982; Douglaston, N.Y.

Floyd W. Iden, '41; June 27, 1984; Pompton Plains,N.J.

Robert P. Boyer, '42; December 4, 1984; Manchester, Tenn.

Joseph A. Crutcher, '42; April 1985; Fullerton, Calif

Wesley R. Floyd, '42; February 29, 1984; Bradenton,Fla.

Ray O. Wyland, Jr., '42; April 1, 1985; Tujunga,Calif.

Margaret L. Blizard, '43; May 10, 1985; Norwood.Mass.

Clyde A. Booker, Jr., '43; 1985; Pittsburgh, Penn.

Shao T. Hsu, '43; 1985; Bethesda, Md.

James B. Hoaglund, '45; February 9, 1985; Minne

apolis, Minn.

David A. Trageser. '45; April 26, 1985; Wayland.Mass.

William H. Auerswald, '46; April 10, 1985; Suffield,Conn.

Jacob W. Ullmann, '46; February 5.1985; New York,N.Y.

Helmar Schlein, '47; February 19, 1985; Reseda,

Calif.

John F. Christopher, '48; April 24, 1985; St. Louis,

Mo.

Ralph Segal, '48; January 1985; Roslyn, N.Y.

David S. Selengut, '48; July 24, 1984; Schenectady,N.Y.

Hcndrie J. Grant, '49; August 1, 1984; St. Paul,

Minn.

Leonard N. McKibben, '49; December 31, 1984;

Westlake, Ohio.

Waller E. Mutter, '49; November 1984; Poughkeep-sie, N.Y.

Mctr Drubin, '50; April 25, 1985; Rancho Palos

Verde, Calif.

Robert S. Lovett, '50; June 1984; Wilmington, Del.

Dean E. Cogswell, '51; May 11, 1985; Wenham,

Mass.

Bryant W. Foster, '51; March 17, 1985; Braintree,

Mass.

Charles G. Etler, Jr., '53; December 22, 1983;

Wayne, Penn.

Lawrence Tsun-Ying Kwan, '53; December 1967;

Encino, Calif.

Alan K. MacKenzie, '54; April 10, 1985.

Tse H. Chu, '55; March 1982.

Agnes M. Galligan, '55; 1985; West Roxbury, Mass.

Jakob Auslaender, '56; February 5, 1985; Chicago,

III.

Waller C. Kottemann, '59; Mav 5. 1984. Clarendon

Hills. 111.

Vinod Sundra, '59; December 20, 1980; Bedford,

Mass.

Michael Vincent Bogda, '66; May 22, 1983; Wood

land Hills, Calif.

Paul G. Maguire, 68; April 4. 1984; Aclon, Mass.

Theodore R. Sieger, Jr., '74; 1985; Longmeadow.

Mass

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TECHNOLOGY REVIEW A29

lution announced above provides the only male in

Iwo.

Also solved by Jeffrey Mattox, John Bobbin, Mat

thew Fountain, and Neil Hochstedler.

APR 2. Note that each of the following is a different

way of evaluating the same equation:

16601.92 + 14374.08 =

11334.4 + 19641.6 =

18521.44 + 12454.56 =

4147.36 + 26828.64 =

Write the equation in the usual form.

John Bobbitt was able to recognize the hidden

quadratic and writes:

The four "sets of numbers" represent

a! + 2ab + b! = (a + b)1where a = 111.6 and b a 64.4. By forming different

groupings of the four terms a!, b2, ab, ab, we cangel the four sets. Thus,

(a2 + b!) + 2ab = (a + b)!(ab + b2) + (ab + a1) = (a + b)2

(b2 + 2ab ) + a2 «■ (a + b)2b2 + (a2 + 2ab ) = (a + b)2

Also solved by Harry Zaremba, Jerry Grossman,

Marshall Fritz, Matthew Fountain, and Winslow

Hartford.

APR 3. Find all maxima and minima of

f)In/1 + e" +

m(i + e- - I)

without using calculus.

The solution below is from George Bird:

Note that the expression (1 + e") may be written

as(l + 1/e"). Use the second form of this expression

and write it as a fraction with the common denom

inator e", thus:

(e1 + l)/e*

If we substitute this form, the expression

ln(I + e"*) becomes:

ln(l 4 e") - ln(e»).

The function

|ln(l + e"") + x/2|/|ln(l + e") - x/2]

then becomes:

|ln(e' + 1) - ln(e') + x/21 / [ln(l + el) - x/2|,

which reduces to 1/1. Therefore there are infinitely

many maxima and minima, and all have value 1.

Also solved by Allen Tracht, Edwin McMillan,

Harry Zaremba, Howard Stern, J. Richard Swen-

son. Jerry Grossman, John Bobbin, John Trussing,

Ken Haruta, Marshall Fritz, Matthew Fountain,

Mike Hennessey, Naomi Markovitz, Peter Card,

Ross Hoffman, Steve Feldman, Steve Silberberg,

Tony Trojanowski, Winslow Hartford, and the pro

poser. Rick Decker.

APR 4. On each day of the year (not leap year) you

are given a penny. On December 31 you are given

your last penny and told that it was fresh from the

U.S. Mint, but that one of the previous pennies may

have been counterfeit, and therefore lighter or heav

ier than the standard penny. You are asked to de

termine the number of balancings, using a common

pan balance, that would be necessary and sufficient

to determine whether or not there is a counterfeit

coin, and if there is, to tell whether it is heavier or

lighter than the last penny that you received.

The following solution is from Leon Tabak:

Two balancings are necessary (clearly) and suffi

cient for determining which of three possible so

lutions is the true situation:

(1) all coins are genuine.

(2) one coin is counterfeit and it is lighter that all of

the other (genuine) coins.

(3) one coin is counterfeit and it is heavier that all

of the other (genuine) coins.

Divide the set of 364 coins (whose genuineness is

not known) into three sets: SI, S2, and S3. Let SI

contain 122 coins. Let S2 and S3 each contain 121

coins. A fourth set, G, contains the one certified,

genuine coin.

First measurement: Place SI in the left pan of the

scale. Place S2 and G on the right side of the scale.

If the Iwo balance, then all of the coins in SI and

S2 must be genuine and the counterfeit coin, if there

is one, must be in S3. The possibilities that remain

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if the two pans balance in this first measurementare:

(1) all coins arc genuine.

(2) S3 contains an extra-light counterfeit coin.

(3) S3 contains an extra-heavy counterfeit coin.If the right pan comes to resl at a lower level than

the left pan, then two possibilities remain:

(1) SI contains an extra-light counterfeit coin.

(2) S2 contains an extra-heavy counterfeit coin.In either case, all of the coins in S3 must be genuine.

(There is at most one counterfeit coin, and this lesthas shown lhat it must be in SI or S2).

Second measurement: Place SI in Ihe left pan. Place

S3 and C in Ihe right pan If Ihe scale balances on

the second measurement and it also balanced on

the first measurement, then all coins are genuine.

If the left pan is higher, but the scale balanced dur

ing Ihe first measurement, then S3 must contain an

extra-heavy counterfeit coin. If the left pan is lower,

but the scale balanced during Ihe first measure

ment, Ihen S3 must contain an cxira-light counter

feit coin. If the scale balances on the second

measurement, but the left pan rested lower during

the first measurement, then S2 must contain an ex

tra-light counterfeit coin. If the scale balances on

the second measurement, but the left pan rested

higher during the first measurement, then S2 must

contain an extra-heavy counterfeit coin. If the left

pan is lower and the left pan was also lower during

the first measurement, then SI must contain an ex

tra-heavy counterfeit coin. If the left pan is higher

and the left pan was also higher during the first

measurement, then SI must contain an extra-light

counterfeit coin. Two outcomes are inconsistent

with Ihe statement of the problem. The left pan

cannot be higher in the first measurement and lower

M.I.T. ALUMNI

CAREER SERVICES

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of jobs for alumni across

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We want more firms to

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We want more alumni mak

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Whether you have a |ob to

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let us send you a copy of

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Call or write

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Alumni Career Services

M.I.T., Room 12-170

Cambridge, Mass 02139

Tel: (617) 253-4735

in the second, nor can it be lower in Ihe first and

higher in the second. Either outcome would imply

the existence o( more than one counterfeit coin.

Also solved by Rik Anderson. Dudley Church,

E.P. Schacht, Frederic Jelen, Harry Zaremba, John

Prussing, John Spalding. Kenneth Olshansky. Mat

thew Fountain, Phelps Meaker, Walter Guett, and

the proposer, Alan Faller. Mr. Faller also showed

that with six balancings, one can also determine

which coin was counterfeit (except for leap years

when seven are required). Copies of Mr. Fa'ller'ssolution arc available upon request.

APR 5. Find X for Iwo configurations—when A ^

15, B = 10, and. H = 8; and when A " 16 + 2V2.

B = 16 - 2V2, and H = 2.

Howard Stem's solution is one of the few in

which an exact solution was found for the second

configuration:

With the above lengths labelled, similar triangles

give the following relationships:

(VB\ - X'VX = H/Y

(VA! - X2)'X = H/(X - Y).

Eliminating Y yields:

1/(V'B; - X;) - 1/(VAJ - X2) = 1/H (1)

The left side of (1), viewed as a function of X, has

a minimum of (l'A) + (!/B) when X = 0 and in

creases with X. For the first set of parameters given

(A = 15, B = 10, H = 8), the left hand side of (I)

will always be greater than 1/6 = 1/15 + 1/10. But

the right hand side is 1/8. Therefore, (1) can never

be satisfied; so these "crossed ladders" represent

an impossible configuration. The highest point of

crossing is 6, never 8.

The second set of parameters do allow for a phys

ically possible solution. After substituing A = 16

+ 2(2)'2, B = 16 - 2(2)" and H = 2 in (1), thesolution is X = (168)" = 12.96.

Also solved by Allen Tracht, Avi Ornslein, Fred

eric Jelen, Harry Zaremba, Matthew Fountain, Peter

Card, Ross Hoffman, Steve Feldman, Winslow

Hartford, and the proposer, Martin Brock.

Better Late Than Never

1984 N/D 1. Neil Hochstedler notes that the third

move m the third variation should be Q-a4 mate

1985 JAN 3. Randall Whitman has responded.

F/M 3, F/M 4. Ceorge Parks has responded.

APR SD2. E.P. Schact wants to rescind Horton's

Nobel Prize believing that the chicken came first.

Proposer's Soluiions to Speed Problems

SD 1. Zero.

SD 2. Don't cash your fourth heart! Lead a club and

trust that declarer now thinks that it is safe to take

the club finesse into your hand.

♦ K93

V 653

♦ AJ92

♦ Q74A J542

V 10 9 8

♦ 863

A 8653 A AQ8

VJ42

♦ K 10 4 3

A A K3

A 10 7 6

V A KQ7

♦ Q75

A J 10 9

Ml AUGUST/SEPTEMBER 1985

PUZZLE CORNER

ALLAN J. GOTTLIEB

Playing 20 Hands With

No Winners, and the Feeding of Cats

and Explorers

Since this isr the first issue of a new

academic year, I once more review

the ground rules under which this

department is conducted.

In each issue I present five regular

problems (the first of which is chess,

bridge, or computer-related) and two

"speed" problems. Readers are invited

to submit solutions to the regular prob

lems, and three issues later one sub

mitted solution is printed for each

problem; I also list other readers whose

solutions were successful. For example,

solutions to the problems you see below

will appear in the February/March issue.

Since I must submit that column some

time in November (today is July 19), you

should send your solutions to me during

the next few weeks. Late solutions, as

well as comments on published solu

tions, are acknowledged in the section

"Better Late Than Never" in subsequent

issues.

For "speed" problems the procedure

is quite different. Often whimsical,

these problems should not be taken too

seriously. If the proposer submits a so

lution with the problem, that solution

appears at the end of the same column

in which the problem is published. For

example, the solutions to this issue's

"speed" problems are given below.

Only rarely are comments on "speed"

problems published or acknowledged.

There is also an annual problem, pub

lished in the first issue of each new year;

and sometimes I go back into history to

republish problems that remained un

solved after their first appearance.

Problems

OCT1. We begin with a bridge problem

from Lawrence C. Kells, who writes that

a friend directed a duplicate bridge tour

nament in which every hand was played

20 times. One of these hands produced

a strange result. At four of the tables the

final contract was one club, played once

from each of the foursides. At four of

the other tables it was played at one dia

mond once from each side. At the re

maining tables it was played once from

each side at one heart, one spade, and

one no-trump. Every one of these con

tracts was set. Analysis of the hand

proved that none of the declarers made

a mistake in play. Unfortunately, Mr.

Kells failed to see what the deal was.

Can you reconstruct it? (That is, a deal

where any contract anybody bids can be

set no matter how hard he tries to make

it.)

OCT 2. Smith D. Turner (Jdt) has a

question about the following nim-like

game, called the game of thirty-one, that

was popular with New York City ma

gicians about 40 years ago:

Twenty-four cards, the 1-6 of each suit,

are put face-up on the table. Two players

pick up alternately, each keeping track

of his pip-total. One wins by hitting

thirty-one, or forcing one's opponent to

exceed 31. Does the first or second

player have a sure win, and how does

he play to insure it?

OCT 3. Albert Mullin asks us an edu

cational problem:

A very wise dean wishes to improve the

quality of research and education at her

university. She believes some of the

"fault" is with the department heads,

but, being wise, recognizes some of the

"fault" may be with herself. For starters,

she decides to relocate her office so as

to minimize the average distance to her

department heads. Where should she

locate her office? (For simplicity, assume

all offices are located in a plane and that

the dean cannot co-reside with any de

partment head).

OCT 4. Yogesh Gupta posted the fol

lowing problem on an electronic bulletin

board that I read:

An exploring team wants to reach a des

tination that is six days away. Each ex

plorer can carry enough provisions to

sustain one person for four days (and

the distance an explorer travels in a day

is independent of the amount of provi

sions he or she is carrying). What is the

smallest team that permits at least one

explorer to reach the destination and

permits all the explorers to return home

safely?

OCT 5. John Glenn has two cats, which

are fed a single can of cat food. The food

comes out of the can as a single intact

cylinder, and the can is then used to cut

the cylinder into two shares of food.

Where should the can be positioned to

generate equal shares?

Food

Can

Speed Department

SD 1. Phelps Meaker wants to know at

what time between two and three o'

clock the hour and minute hands are in

a symmetrical position.

SD 2. We end with a series of "chem

istry" questions from The Tech, M.I.T.'s

student newspaper:

Give the names of the following com

pounds:

I WoNoCrKrWsNoCrKrVWNoCfKrWoNeCrKrWlNoCrierWoNoCtKl

2. Be +At-

3. BotuHIJKLMnO

4.

10 TOR TOR

9. (BoNa2)l2

F.++

A30 OCTOBER 198S

Melvin Moisted, '23; May 12, 1985; Philadelphia,Penn.

Millon O. Orwin, '23; December 11, 1984; Kala-mazoo, Mich.

Lewis J. Powers, '23; 1985; Springfield, Mass.

Mrs. Alice C. Bishop, '25; Januarv 20, 1982; Cran-ford, N.J.

Kathlyn A. Carnagey, '25; October 7, 1983; Badin,N.C.

William J. Mahoney, '25; June 12, 1985; Chatham,Mass.

Vincent B. Bennett, '26; March 1985; Ipswich, Mass.

Philip A". Hendee, '26; September 25,1984; Charleston, W.V.

William H. Hoar, '26; April 27, 1985; SUver SpringMd. V BJames S. Offuti, '26; May 14, 1985; Seminole, Fla.Alfred P. Steensen, '26; March 30, 1985; Manchester, N.H.

Mrs. Edward). Poitras, '28; February 16,1985, VeroBeach, Fla.

John L. Cantwell, '29; May 8, 1985; Chicopee Falls,Mass.

Willis F. Davis, '29; April 14, 1985; Erie, Penn.Harold C. Pease, '29; May 30, 1985; St. Petersburg,

John P. Rich, '29; November 22, 1984; Nashua,N.H.

Warren W. Walker, '29; April 20, 1985; Montclair,N.J.

John F. Guinan, '30; May 18,1985; Arlington, Mass.Bertwell M. Whitten, '30; June 5, 1985; Searsport,Maine.

Alexander Cameron Crdsman, '32; 1985; Santa Ana,Calif.

Lawrence W. Grady, '32; June 6, 1985; South Yarmouth, Mass.

Samuel E. Paul, '32; May 4, 1985; Napa, Calif.

William D. Murphy, '33; May 22,1985; Annandale,Va.

John B. Smyth, '33; April 3, 1985; Peekskill, N.Y.Charles H. MacFarland III, '34; June 13, 1985;Union, Conn.

Herbert A. Morriss, Jr., '34; February 9, 1985; Manhattan Beach, Calif.

Gerald M. Reed, Jr.,'34; July 4,1985; North Quincy,Mass.

Alfred Z. Boyajian, '35; 1985; Scituate, Mass.Willard R. Crout, '35; October 1983; Pittsburgh,Penn.

Alex G. Keiller, '35; 1985; Encinitas, Calif.

Alwin R. Knoeppel, '35; April 1984; New York,N.Y.

Robert T. Sutherland, Jr., '35; May 31, 1985; SandyHook, Conn.

Lincoln P. Vennard, '35; November 1984; Burbank,Calif.

Richard M. Whitmore, '35; March 3,1985; Whiting,N.J. 6Dana Devereux, '36; May 15, 1985; New Canaan,Conn.

Robert L. Alder, '37; April 6, 1985; Garland, Tex.

Melville E. Hitchcock, '37; March 9,1985; Ivoryton,Conn.

Leslie A. Johnson, '37; April 20,1985; Sanford, N.C.

Wayne M. Pierce, Jr., '37; May 7, 1985; Orange,Conn.

George Albert Randall, '37; June 17, 1985; New-buryport, Mass.

John A. Dodge, '39; May 1985; Wayne, Penn.

William G. Hamlin, '42; June 20, 1985; Glenshaw,Penn.

Everett L. Meley, Jr., '42; April 25, 1985; Houston,Tex.

Aurelio C. Hevia, '43; May 29, 1985; Miami, Fla.Martin H. Winter, '43; February 10,1985; Paramus,N.J.

Stanley Berinsky, '44; March 4, 1985; Sunnyvale,Calif.

Frederick B. Meier, '44; 1985; Monroe, Mich.

Clarence S. Howell, Jr., '45; June 7, 1985, MercerIsland, Wash.

Donal Botway, '49; May 8, 1985.

Louis Robinson, '50; March 28, 1985; Scarsdale,N.Y.

Roland F. Beers, Jr., '51; March 20, 1985; Dorset,Vt.

Robert W. Elliott, '52; May 13, 1985; LynnfieicMass.

Morton S. Hoppenfeld, '52; March 26, 1985.

I,

Shanker R. Bagade, '55; January 1979; Cincinnati,Ohio.

Choji Nozaki, '55; April 16, 1985; Nagoya, Japan.

Leon A. Yacoubian, '55; April 1985; Damascus,Syria.

Dalton L. Baugh, Sr., '56; January 3, 1985; Malta-pan, Mass.

Charles Diebold III, '58; June 1985; Buffalo, N.Y.

Myron P. Lepie, '58; May 9, 1985; Chestnut Hill,Mass.

Raymond A. Bruce, '61; April 13, 1985; Middle-

town, N.J.

Charles C. Conley, '61; November 20, 1985; SaintPaul, Minn.

Sudhanshu K. Dikshit, '61; October 1982.

Clarke E. Swannack, '61; December 12,1984; Beloit,Wise.

Albert L. Baker, Jr., '65; March 1985; Dunedin, Fla.

Vittorio Baldini, '75; April 23, 1985; Milano, Italy.

John B. Lamb, '77; May 31, 1985; Wellesley, Mass.

David E. Miller, 79; July 1, 1985; Edwards AFB,Calif.

David J. Shapiro, '87; June 19,1985; Concord, Mass.

StemhiecherGup.


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Specializing In the

solution of complete.

difficult, now,

Infrequent, or unique

problems

TECHNOLOGY REVIEW A2»

Solutions

M/J 1. How does South make five clubs after an

opening lead of the »Q, and what lead sets this

contract?

♦ A 84

¥ AK 10

♦ A7654

♦ KJ

A K

V QJ9432

♦ Q9832

+ Q

* QJ10 3 2

V 865

♦ JA 107 54

♦ 9765

V 7

♦ K 10

♦ A98632

Benjamin Feinswog suggests winning the opening

lead with the ♦ K and then cashing the AK and

*J, discarding a spade. Now the V10 is ruffed and

the *A cashed. At this point the ♦ 10 is led and,

whether East ruffs or not, the ♦A provides an entry

to the ♦ A permitting a second spade discard. ThusSouth loses only one spade and one trump. Mr.

Feinswog notes that a lead of the ♦ 2 (among others)

will set the contract.

Also solved by Richard Hess (who believes that

any lead will set the contract), Matthew Fountain,

and the proposer, Doug Van Tatter.

M/J 2. A large bed of flowers and greenery is laid

out in the form of a regular polygon of N sides. A

walk composed of N trapezoidal concrete slabs sur

rounds the flower bed. A circumscribing circle pass

ing through the outer corners of the walk and an

inscribing circle tangent to the inner flats of the

trapezoids have circumferences in the ratio of al

most exactly 191:165. The total area within the outer

periphery of the walk and the area of the walk itself

are in the ratio of 4:1. Find N.

The following solution is from Richard Helden-fels:

The general equations of regular polygons and a

trigonometric relationship were all that was re

quired to obtain a direct solution. The ratio of cir

cumscribing and inscribing circles provides thefollowing:

a.tan(180/N)/a1sin(180/N) = 191/165,

where the a's are the length of the outer and inner

sides and N is the number of sides. The rano ofareas provides the following:

A./Ao = (a,/a<,)J = 3/4.

These equations can be combined and simplified toyield:

cos(180/N) = (165/191)(2/V3).

The solution is: 180/N = 4.0391°. Since the ratio ofcircles was not exact, N = 45. The ratio of 191:165

is greater than that ratio for N = 45 polygon by the

factor of 1.000048. A drawing of one trapezoidalconcrete slab follows:

26646

25 526

23 076

Also solved by Charles Freeman, Dennis White,

Frank Carbin. George Parks.Harry Zaremba. John

Prussing, John Woolston, Matthew Fountain, Mi

chael Jung. Naomi Markovitz, Richard Hess, Steve

Fcldman, and the proposer, Phelps Meaker.

M/J 3. Given a number N between 500 and 1000,

rapidly construct a scries of numbers using eight of

the nine digits from 1 through 9 once, the other

digit twice, and a few zeroes so the series totals N.

Naomi Markovitz must have spent a great deal

of time in tricky parlors:

HTU

1

2

3

4

5

6

7

8

9

HTU

1

2

3

4

5

6

7

8

9

H T 11

1

2

3

4

5

6

7

8

9

HTU

1

2

3

4

5

6

7

8

9

3

Being that as many zeros as desired may be used,

I'll worry about whether the other digits should be

in the hundreds, lens, or units column and then

affix the appropriate zeros in the presentation of the

requested series. 1 noticed in the example presented

that most of the non-zero digits are in the tens column. Indeed the sum of 10 + 20 + . . . + 90 =

450, which is the same order of magnitude as any

answer which may be requested. Therefore, my ap

proach would be to start with one "copy" of each

digit in the tens column and then to check, in the

case of each sum desired, which digits have to be

moved to the hundreds column, which to the units

column, and which has to be duplicated (and the

column in which the second copy will appear). If a

digit, a, moves from the lens to the hundreds col

umn, the sum will increase by 100a - 100a = 90a.

If a digit, b, is moved from the lens column to the

units column, the change is - 10b + b, or a decrease

of 9b. Take, for example, the case which is pre

sented in the problem: N = 642.1 start off with the

arrangement at the left of the box above. This gives

a total of 450. 1 have to add 642 - 450 = 192. To

check which digit has to be moved to the hundreds

column, I divide 190 by 90 and get 2 +. I'll move 3

to the hundreds column, because if 1 add too much,

1 can subtract the extra by moving a digit to the

units column. Now I have the second display in the

box. I've added 3 x 90 = 270, instead of the re

quired 192, so I'll have to subtract 78. To determine

which digit to move to the units column, I divide

78 by 9 and get 8 +. Again, I chose to subtract "too

much" because this problem can be solved by add

ing the second copy of one of the digits. So I'll move

the 9 into the units column. Now I have the third

display in the box. I have subtracted 9 x 9 - 81

instead of the required 78. I have to add 81 - 78

= 3. I therefore add a second 3 to the units in the

fourth display. So, the answer can be presented as:

319

20

40

50

60

70

63

642

There are, however, cases in which the procedure

must be slightly modified. The basic procedure will

work in all cases since:

500 « N « 1000

we see that

50 « N - 450 « 550.

Dividing by 90 will always give an, answer in which

a single digit must be moved to the hundreds col

umn. (Even if the division comes out even, add 1

to the quotient to determine the digit to be moved,

because eventually a second copy of one digit will

have to be added.) The second stage consists of

dividing by 9 a number from 1 to 90. Several prob

lems could occur:

(1) The number that has to be moved from the tens

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Evaluation,

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Alexander Kusko '44

161 Highland Avenuo

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02194

(617)444-1381

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J. D. Okun. 75

K. J. 0 Reilly. '80

R. M. Simon. 72

T. von Rosenvinge IV.

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W. G. Williams. '65

0. W. Wood. 76

column to the units column is no longer in the tens

column, because it has already been moved to Ihe

hundreds column. Solution: If b £ 3, move both 1

and b - 1 (or any other combination of 2 numbers

totaling b) into Ihe units column. If b - 2, instead

of moving 2 into Ihe hundreds column (for a gain

of 2 x 90 = 180). move the 3 into the hundreds

column and then 1 + 9 (or any other combination

totaling 10 other than 2 + 8) into the units column:

3x 90 -1x9-9x9 = 180

The 2 will then be available to move into Ihe units

column.

If b " 1, instead of moving 1 into the hundreds

column, move 2 inlo the hundreds column and

either 3 and 7 or 4 and 6 into Ihe units column. The

1 will then be available to be moved into the units

column.

(2) It's required to move 10 or 11 into the units

column. Solution: Move in two numbers with the

appropriate sum. There will always be a pair avail

able, because the original list (1-9) consists of four

pairs totaling 10 and four pairs totaling 11. Since

only one digit has been removed from the tens col

umn,enough choices still remain. The last stage al

ways gives a number from 1 to 9 that can be added,

so this can be accomplished by placing the appro

priate single digit in the units column.

Also solved by Mathew Fountain, Richard Hess,

Steve Fcldman, and the proposer. Lester Steffens.

M/J 4. As we generate geometric figures to repre

sent y ■=> x", we have "elements" consisting of

points, lines, faces, cubes, etc. as n increases. For

the number of points in each figure, we have (for

n > 0) P = 2". Derive the number of lines and faces

in a five-dimensional hyper-cube.

Charles Sutton solved a generalization of M/J 4:

We may consider the generation of the sequence of

n-dimensional cubes, d units on a side, as follows:

A cube of any number of dimensions, moved a dis

tance d in a direction perpendicular to all its di

mensions, will generate a cube of one higher

dimension. If one starts with a zero-dimensional

cube (a point), this procedure will generate, in

succession, line segments, squares, cubes, and so

on into cubes of higher dimensions (hypercubes).

Nole that an n-dimensional cube will have as com

ponents all cubes of lower dimensions, from zero-

dimensional points up to (n - l)-dimensional

cubes. Now when an n-dimensional cube is moved

a distance d in Ihe direction of the next higher di

mension to generate an (n + l)-dimensional cube,

its r-dimensional component cubes (0 s i < n) will

necessarily generate (r + l)-dimensional compo

nents of Ihe (n + l)Klimensional cube. It is clear

that Ihe number of points will double, since there

will be just as many points in the final position of

the cube as in Ihe initial position, and hence the

number of points (zero-dimensional components)

in an n-dimensional cube will be 2". When an r-

dimcnsional component (r s> 1) of an n-dimensional

cube moves a distance of d, there will be just as

many such components in Ihe final position as in

Ihe initial position, but there will be additional r-

dimensional components generated by the motion

of (r - l)-dimensional components. Hence the

number of r-dimensional components in an (n +

l)-dimensional cube can be obtained by adding the

number of (r - l)-dimensional components to twice

Ihe number of r-dimensional components in an n-

dimensional cube. We can use this to complete the

array at the top of the next column, giving the lower

dimensional components for cubes of each number

of dimensions. Filling in l's in the diagonal (since

an n-dimensional cube can be considered to have

itself as a component) and powers of two in the left

hand column, we can continue by adding twice the

value of any entry to its left hand neighbor and

writing the result below the doubled entry. The con

struction of this array is reminiscent of Pascal's tri

angle, and in fact the numbers in the rows can be

seen to be the coefficients in the expansion of (2x

+ 1)". And the answer to Winslow Hartford's ques

tion in M/J4 is that the numbers of lines and faces

in a five-dimensional hypercube are both 80.

Mr. Sutton then adds the following remarks:

I have always been intrigued by the geometry of

Dlncnslon of components

2 3 4 5 6 7

Dloenslon

of cube

0;

2

3

4

s

6

r

l

2

4

8

16

32

64

128

1

4

12

32

80

192

448

1

6

24

90

!40

672

1

8

40

160

160

1

10

160

290

1

12

84

higher dimensional figures, and recall having read

someplace that there are six regular four-dimen

sional solids but that only the analogues of the cube

and tetrahedron exist for all higher dimensions.

That set me to wondering what the triangular array

I had obtained in my solution of M/) 4 would look

like for the equilateral triangle, regular tetrahedron

scries. Actually it's fairly easy, since to step up one

dimension you need only to find a point in the next

higher dimension that is the same distance from Ihe

set of equidistant points in the lower dimensions.

The n-dimensional analogue of Ihe tetrahedron

would have n + 1 vertices, and the number of its

r-dimensional components (each having r + 1 ver

tices) would be the number of combinations of n +

1 things taken r + 1 at a time. Binomial coefficients!

And the numbers in the horizontal row of the tri

angular array corresponding lo the n-dimension.il

analogue of the tetrahedron turn out to be the co

efficients of (x + 1)"'' - x"*1. As a check, for n =

3, a tetrahedron has four vertices, six edges, four

faces, and one tetrahedron, coefficients of (x + If*1- x"*1.

Also solved by Avi Omstein, Charles Freeman,

Dennis White, Matthew Fountain, Richard Hess,

and Winslow Hartford.

M/J 5. Take the letters in the first half of alphabet

in order. Place the A on the table. Place the B next

to the one of the four sides of the A. Place the C

next to one of the six sides at the AB (or BA) pair.

Then add the D and so on. If you do this correctly,

when you reach the letter M you will have created

a crossword-puzzle matrix of complete common

English words, no proper names, no foreign words,

and no acronyms or abbreviations. Having solved

the problem as posed, can you add one or more

letters lo the A-M set and still retain complete

words?

Both Rik Anderson and Matthew Fountain found

the identical solution—one that goes up to P. The

key to extending past M is the word "knop." It

would be hard to go further since adding Q would

require U and hence R S T.

H

BACK

FED N

JIG

L

H

0

P

Also solved by Harry Zaremba, M. Arch, Naomi

Markovitz, Richard Hess, and Steve Fcldman.

Proposers' Solutions to Speed Problems

SD1. 2:46 2/11

SD2.

1) Polywannacracker

2) Polar bear

3) Barium Coldwater

(barium gold H-lo-O)

4) Hi-O, Silver

5) A dozen bananas

6) Paramedics

(or paradox)

7) Metaphysics

8) Orthodox

9) Methyl ethyl

chicken wire

10) Transistor

11) Cis-boom-bah

12) FORTRAN 4

13) Mercedes benzene

14) Ferrous wheel

SEND PROBLEMS. SOLUTIONS.


COTTUEB. -67. ASSOCIATE




NEW YORK UNIVERSITY. 2S1


N.Y.. 10012.

A32 OCTOBER 1985

PUZZLE CORNER

ALLAN J. GOTTLIEB

Where Squares

and Office Parties Mix

Jim Landau sent me quite an unusual

package. I have often received mail

marked fragile, but never before one

mailed indestructible. Inside was a rope

tied into a special knot and an accom

panying problem. Although I do not as

yet see how to present the problem

without having each issue of Technology

Review include an attached rope, I have

had fun with the problem myself. I

should also remark that this indestruc

tible parcel seemed to get considerably

better treatment en route than many un

marked packages I have received. Per

haps stores should try sending crystal

in packages marked indestructible.

On a personal note, I am pleased to

report that I have received tenure at

NYU.

Problems

N/D 1. We begin with a computer-re

lated problem from Al Weiss, who is

given the coordinates of four points and

wants to determine if they form the four

corners of a square. No assumption can

be made about the order in which the

points are presented. For example, a

square might occur as:

(8,8), (3,5), (2,4), (4,9).

Mr. Weiss seeks not just a solution, but

an elegant algorithm.

N/D 2. Oren Cheyette offers us a sea

sonal problem. For an office party, each

person is supposed to bring a gift for

someone else. The recipients are as

signed to givers by writing each person's

name on a slip of paper, putting the slips

in a hat, and having everyone draw a

slip. Obviously, it's no fun if someone

draws his own name. What is the prob

ability that in an office of n people, no

one draws his own name?

SEND PROBLEMS, SOLU

TIONS, AND COMMENTS TO

ALLAN I. GOTTLIEB. '67. THE

COURANT INSTITUTE, NEW

YORK UNIVERSITY, 251 MER

CER ST., NEW YORK, N.Y.

10012.

(notN/D 3. Given a square matrix A

necessarily invertible), satisfying

AA = AA',

where the prime signifies transpose op

erator, Howard Stern wants you to show

that

A = A'

using matrix operation only, i.e. without

using normed algebras and approxi

mating A by an invertible matrix.

N/D 4. Walter S. Cluett asks: What is the

lowest number of current U.S. coins (1

cent through $1.00) for which there is

no combination of coins that will equal

in value a single coin? How many such

quantities are there under 100?

N/D 5. Ronald Raines wants you to find

functions f and g satisfying

f[f(x)] = x

glgMl = - xfor all real values of x.

Speed Department

SD 1. Phelps Meaker has a pan with

perpendicular ends and sloping sides. It

is two inches deep and measures 8" x

10" on the bottom. What is the slant

height of the sides if the capacity is 200

cu. in.?

SD 2. Jim Landau wants to buy a solid-

state digital clock with a 12-hour LED

display and wonders at what time

would the largest number of LEDs be

on? The smallest number?

Solutions

JUL1. Find the smallest prime number that contains

all 10 digits.

Dan Schmoker combined analysis with some

computer work to obtain the following solution:

The smallest 10 digit number with all digits different

is 1,023,456,789. The sum of the digits in this num

ber is 45 which is divisible by 3; hence the number

is divisible by 3 and is not prime. In addition, any

rearrangement of this 10-digit number would also

be divisible by 3 and not prime. It follows therefore,

that the smallest prime number that contains all

digits must be an 11-digit number. The extra digit

cannot be 0, 3, 6, or 9; otherwise the new number

and any number formed by rearranging the digits

would be divisible by 3. and hence not prime. The

smallest number which contains all digits and could

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possibly be prime is therefore the 11-digit number10,123,456,789. The prime lest should be started

with this number. The numbers to be tested forprime are generated by rearranging the 11-digit

number, minimizing the change in value by switch

ing the right-hand digits first. The prime test is done

by first generating a list of prime numbers. On my

machine with 48K of memory the biggest list of

prime numbers I could generate had a total of 9000

numbers with the highest prime number being

93001. By dividing with this list of prime numbers,the number 10,123,457,689 (the seventh number

considered) was identified as a possible prime num

ber. At this stage 1 began dividing by all odd numbers greater than 93001, continuing until the divisor

exceeded the square root of 10,123.457,689. This last

test only took a few minutes and showed that

10,123,457,689 was indeed prime.

Also solved by Donald Savage, Henry Hirsch-land, Jonathan S.idick, N.C. Strauss, Ned Staples,P.V. Heftier, Richard Hess, and the proposer, Matthew Fountain.

JUL 2. The original Tower of Hanoi, first described

in 1883, consisted of 64 golden disks, each of a dif

ferent diameter, stacked according to size, with the

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smallest on top. The tower is to be restacked on one

of two additional sites, moving one disk al a time

off the top of one stack either to an empty site orto the top of the slack on one of the other sites,

without ever placing a larger disk on a smaller. The

original problem was to find the minimum number

of moves to transfer the enlire stack. The new prob

lem is to calculate the location of each disk after

1,001 moves have been made using the optimum

transfer procedure.

Rik Anderson found it easier to take a gianl step

forward and then 23 small steps back:

Identify the original stack as position A. If disk 1 is

initially moved to position C, Ihen disk 2 will ini

tially be moved to position B al move 2, disk 3 will

go to C al step 4, disk 4 to B al step 8, etc. Disk n

will first move at move 2'" ", to position B if n is

even, to C if n is odd. Following this rule, the first

time disk 11 moves is at move 2'° = 1024, lo position

C. At the previous move, number 1023, disks 1-10

would have reached position B. In getting lo this

position, disk 1 is at any position for two moves,

disk 2 moves every 4th move, disk 3 ever)18th move,

and disk 4 every 16th. Disk l's moves are in reverse

sequence (A,C,B,A,C,B), as are all odd-numbered

disks, while even-numbered disks visit the posi

tions in sequence (A,B,C,A,B,C). Working these

patterns backwards from step 1023, we would find

at slep 1001:

Position A - disks 1, 4, and 11 to 64

Position B • disks 2, 3, and 6 to 10

Position C - disk 5.

Also solved by Dennis While, John Prussing,

Matthew Fountain, Ned Staples, Richard Hess,

Winslow Hartford, and the proposer, Lester Stef-

fens.

JUL 3. Define an n-triangle to be a collection of n(n

+ l)/2 points regularly spaced into the shape of an

equilateral triangle with n points on a side. Define

a 3-line to be a line segment connecting exactly three

adjacent points parallel to a side of an n-triangle.

(The three adjacent points are said to be "covered"

by the 3-line). For what values of n can all points

of an n-triangle be covered by non-intersecting 3-

lines?

This question appears to be difficult, and I con

sider the problem to be still open Harry Zaremba

shows that if one drops Ihe requirement that 3-lines

are parallel to a side of the triangle, then a solution

exists for all

n = 9m and n = 9m - 1.

Dennis White, Richard Hess, and Winslow Hartford

express the belief that (with the parallel require

ment) no solutions are possible. Mr. Hartford notes

thai to have a multiple of three points, we need

n = 3m or n = 3m - 1.

and indicates that an inductive proof should be pos

sible.

JUL 4. A manufacturer makes all possible sizes of

brick-shaped blocks such that the lengths of the

edges arc integral multiples of the unit of length,

and that the number of units in the total length of

the twelve edges of the block is equal lo two-thirds

of the number of units of volume in the block What

sizes does he make?

The following solution is from Howard Stern:

AB:CD x E = FG:H1

Since the smallest value that AB can assume is 12,

it follows that E cannot be larger than 4. It also

cannot be 1 because then the times would be the

same. Therefore, E must be 2, 3 or 4. This forces A

to be 1. In addition, C, F and H must be 2, 3, 4 or

5 for the times to make sense. Since A = 1, E cannot

be 2 because then F would also have lo be 2. E

cannot be 4 either because then B would have to be

2 or 3, and it would be impossible for C, F and H

to be 2, 3 or 5. Therefore, E must be 3. This forces

C, F and H to be 2, 4 or 5, and D and I must be 6,

7, 8 or 9. From Ihe multiplication by 3, the only

possibilities are D = 6 and I = 8, or D = 9, and I

- 7. These restrictions leave only a few feasible

times to try and the only one that works is: 18:49

x 3 = 56:27.

Also solved by Avi Omstein, Dennis White,

Frank Carbin, Harry Zaremba, Henry Hirschland,

Matthew Fountain, P. Michael Jung, P.V. Heftier,

Richard Hess, Steve Feldman, Winslow Hartford,

and the proposer, Phelps Meaker.

JUL 5. An ideal pulley system supports a bucket of

water on one rope and a monkey on the other. The

bucket and monkey are in static equilibrium and at

the same vertical level. Suddenly the monkey began

climbing up its rope. Describe the motion, if any,

of the bucket.

Everyone agrees that the bucket follows the mon

key exactly, independent of the mechanical advan

tage of the pulley system. The argument goes as

follows: The monkey's change of state from rest to

an upwards velocity implies acceleration which re

quires force. The reaction force is tension xT in Ihe

monkey's rope (where x is the pulley's mechanical

advantage with respect to the monkey). The dy

namical response of the bucket to the force xT is

identical to the monkey's, so that the bucket and

the monkey accelerate upwards together. After the

monkey's acceleration stops, it continues to climb

al some velocity v, and so does the bucket.

Solutions were received from Richard Hess (who

attributes the problem to Lewis Carroll), William

Moody (who first heard the problem 58 years ago

in his 8.01 freshman physics course, taught by Prof.

"Hard-Boiled" Lewis Young), Ronald Martin,

Winslow Hartford, Harry Zaremba, Dennis White,

and the proposer, Bruce Calder.


1980 FEB 3. Warren Himmelberger suggests the

solution (33231302212011003)

1983 APR 4. Dick Allphin reports that a similar

problem has appeared in The New York Times. Fur

thermore, Mr. Allphin believes that, when travell

ing in the rain, "for minimum soaking it pays to

make haste."

Y1984 Rik Anderson has noticed that 16 = 41OT.

1984 APR 2. John Colcman has responded.

1985 F/M 3. John Smith found that x = (sin 60)/(sin

75)a.

F/M 4: Phelps Meaker noticed that this problem was

also F/M 4 in 1984. An unintentional yearly problem!

APR 1. Michael Jung has responded.

APR 4. Michael Jung has responded.

APR 5. Frank Carbin has responded.

M/J 4. Jim Landau has responded.

July SD 2. Ben Fcinswog wants to play the VQ

before the 4K and gives a try for the contract, if

West is void in hearts.


SD1. 2(2)5.

SD 2. 10:08 (or 10:08:08 if the clock has seconds).

1:11 (or 1:11:11).

A24 NOVEMBER/DECEMBER 1985

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PUZZLE CORNER

ALLAN J. GOTTLIEB

Great Climbing Monkeys!

This past week I was quite ill with a flu-

like disease that caused me to miss my

first three days of work since arriving at

NYU more than five years ago. And

guess who strolled into the lab to inter

view our research group while I was out:

the CBS Evening News! I guess it just

doesn't pay to get sick so often.

I am sorry to say that, for some unex

plained reason, two of the October so

lutions given in the February/March

issue were unattributed. Somehow be

tween my original manuscript and the

final column, credits to Harry Zaremba

and David Griesedieck for OCT«2 and

OCT 4, respectively, were omitted. I

apologize for the error.

Finally, I would like to acknowledge

a touching letter from one of the most

active contributors to Puzzle Corner

who explained, in a warm and personal

way, why his activity would have to de

crease. This column is dedicated to JohnRule.

Problems

JUL1. For our computer problem of the

month, Matthew Fountain wants you to

find the smallest prime number that

contains at) 10 digits.

JUL 2. Lester Steffens asks us to answer

a new question about the widely known

Tower of Hanoi problem:

The original tower, first described in

1883, consisted of 64 golden discs, each

of a different diameter, stacked accord

ing to size, with the smallest on top. The

tower is to be restacked on one of two

additional sites, moving one disc at a

time off the top of one stack either to an

empty site or to the top of the stack on

one of the other sites, without ever plac

ing a larger disc on a smaller. The orig

inal problem was to find the minimum

number of moves to transfer the entire

stack. The new problem is to calculate

the location of each disc after 1,001

moves have been made using the opti

mum transfer procedure.

JUL 3. Charles Bostick has some points

that need to be covered:

Define an n-triangle to be a collection

of n(n + l)/2 points regularly spaced

into the shape of an equilateral triangle

with n points on a side. Define a 3-line

to be a line segment connecting exactly

three adjacent points parallel to a side

of an n-triangle. (The three adjacent

points are said to be "covered" by the

3-line). For what values of n can all

points of an n-triangle be covered by

non-intersecting 3-lines?

JUL 4. Here is one from a batch John

Rule sent me in 1974 and I have been

periodically milking ever since:

A manufacturer makes all possible

sizes of brick-shaped blocks such that

the lengths of the edges are integral mul

tiples of the unit of length, and that the

number of units in the total length of the

twelve edges of the block is equal to two-

thirds of the number of units of volume

in the block. What sizes does he make?

JUL 5. Here is some monkey business

from Bruce Calder:

An ideal pulley system supports a

bucket of water on one rope and a mon

key on the other. The bucket and mon

key are in static equilibrium and at the

same vertical level. Suddenly the mon

key began climbing up its rope. Describe

the motion, if any, of the bucket.

SEND PROBLEMS. SOLUTIONS.

AND COMMENTS TO ALLAN I

GOTTLIEB. '67. ASSOCIATE



MATHEMATICAL SCIENCES,

NEW YORK UNIVERSITY. 251


N.Y.. 10012.

A22 JULY 1985

Ralph E. Brown, '26; December 24. 2984; Fairless

Hills, Penn.

Robert H. Clarke, "26; November 4. 1985; Dayton.

Ohio.

Thomas J. Eaton, '26; April 17, 1984; Tucson, Ariz.

William M. Smith, '26; 1985; West Hollywood. Fla.

Mrs. William Wraith, Jr., 76; 1983; Tucson. Ariz.

William T. Corey, '27; December 30, 1983; Garden

City, N.Y.

Frederick J. Hooven, '27; February 5, 1985; Nor

wich. Vt.

Roger M. Pierce, Sr., '27; January 10, 1985; West

Brookfield, Mass.

Charles M. Anderson, '28; December 1984; Closter,

Miss.

Albert E. Beitzell, '28; February- 3. 1985; Bangor,

Maine.

Harold A. Harrington, '28; July 16. 1984; Cam

bridge, Mass.

Albert H. Shedd, '28; 1985; Molalla. Ore.

Richard S. Smith, '28; September 24. 1984; Sacra

mento, Calif.

Marshall H. Fay, '29; February 8, 1985; Port Wash

ington, N.Y.

Harold H. Theiss, '29; 1983.

Edward D. Thomas, '29; February 1, 1985; Foxboro,

Mass.

Claude C. Cash, '30; July 25, 1984; Hohokus, N.J.

Richard B. Ellis, '30; March 5, 1985; Athol, Mass.

John C. Larkin, '30; February 3, 1985.

Fred L. Markham, '30; 1985; Provo, Utah.

William Wallace McDowell, '30; March 2.1985; Na

ples. Fla.

Myron T. Smith, '30; February 18. 1985; South

Casco, Maine.

William, E. Yelland, '30; January- 27. 1985; South-

borough, Mass.

Paul A. Davis, '31; December 1, 1983; Vero Beach,

Fla.

Herman H. Ferre, '31; February 1985; Ponce, PR.

Robert M. Kelly, '31; January 30, 1985; Orleans,

Mass.

Manuel Schivek, '31; July 31. 1984; Randolph,

Mass.

Charles R. Wood, '31; September 9. 1984; Kennett

Sq., Penn.

Earl F. Anderton, '32; November 27,1984; Bellevue.

Wash.

Mrs. Amy V. Higgins, Jr., '32; 1982.

G. Edward Nealand, '32; March 19,1985; Sandwich.

Mass.

George W. Palmer, '32; December 20, 1984; Fal-

mouth Foreside, Maine.

Sterling N. Slockbower, '32; 1985; South Plainfield,

N.J.

Walter R. Duncan, '33; January 20,1985; Rosemont,

Penn.

Henry E. Kiley, '33; September 1984; Chatham, N.J.

Harold H. Okasaki, '33; 1984; Gardena, Calif.

John G. Trump, '33; February 21, 1985; Winchester,

Mass.

Arthur E. Byerlein, '34; 1982; Tucson, Ariz.

Robert M. Elliott, '34; December 14, 1984; Norwich,

Conn.

David L. Foulkes, '34; January 23, 1985; Oakland,

Calif.

Alton C. Garland, '34; January 1, 1985; Sandwich.

Mass.

Horace A. Giddings, '34; December II. 1984; Day-

tona Beach, Fla.

James R. Higgins, '34; October 18. 1984; Kansas

City, Mo.

William W. Buechner, '35; March 12. 1985; Arling

ton, Mass.

George R. Bull, Jr., '35; December 9,1984; Wayne,

Penn.

Luis A. Dastas, '35; 1985; Miami, Fla

Richard K. Anderson, '36; 1985; Sumter, S.C.

Charles L. Austin, '36; 1984; Sun City, Ariz.

William Fingerle, Jr., '36; December 6, 1984; Old

Greenwich, Conn.

Carlyle W. Jacob, '36; February 18, 1985; Quincy,

Mass.

George B. Payne,. '36; June 28, 1984; Badiff. Tex.

G. Elliott Robinson, '36; March 30. 1985; Hanover.

Mass.

Annis G. Asaff, '37; March 1983; Lincoln, Mass

Robert L. Carlisle, '37; 1985; Bayside, N.Y.

Peter Kolupaev, '37; January 24,1983; Philadelphia.

Abraham Schwartz, '37; February 4. 1984; Engle-

wood, N.J.

Arthur B. Savel, '38; January 10, 1985; Brookline,

Mass.

Peter E. Kyle, '39; December 6, 1984; Northfield,

Vt.

Parks R. Toolin, '39; 1985; Pittsburgh, Penn.

William C. Walker, '39; January 27, 1985; Lexing

ton, Mass.

Bernard Carver, '40; 1985; Winlhrop, Mass.

Nils M. Rosenberg, '40; January 20, 1985; Seattle,

Wash.

Louis Strymish, '40; 1985; Newton Center, Mass.

Nicholas Williamson, '40; January 20, 1985; Pocas-

set. Mass.

Preston R. Glading, '41; December 16, 1984; Bar-

rington, R.I.

Peter Homack, '41; December 1984; Palm Beach, Fla.

Samuel L. Solar, '41; June 1984; San Jose, Calif.

Robert W. Curtis, '42; November 27, 1984; Arling

ton, Va.

David F. Kinert, '42; 1985; Burlingame, Calif.

John W. McNall, '42; July 10, 1984; Greensburg,

Penn.

Maynard D. Lee, '44; January 3, 1985; Kittery,

Maine.

Leonard T. Loforese, '44; December 13, 1984;

Greenwich, Conn.

Hugh M. Jansen, Jr., '45; February 25, 1984; At

lanta, Ga.

Edward V. Oxenford, '45; February 4, 1985; Buenos

Aires, Argentina.

Edward J. Fradkin, '46; February 10. 1985; New

York, N.Y.

Frederick M. MacDonald, '46; January 8, 1985;

Hyannis, Mass.

Walter H. Amadon, '48; October 26,1984; Sudbury,

Mass.

James Dugundji, '48; January 8,1985; Los Angeles,

Calif.

Peter W. Johnson, '48; 1985; San Jose, Calif.

Mrs. Paul Dulaney, '49; January 1985; Glade Spring,

Va.

William Haddon, Jr., '49; March 4, 1985; Bethesda,

Md.

Worley B. Lynn, '49; 1984; Vian, Okla.

Eugene A. Morgan, '49; March 14, 1985; Bedford,

N.H.

Axel Erik Nygrcn, '51; January 8, 1985; Fagersta,

Sweden.

Sergej Zezulin, '52; February 2.1985; Sea Cliff, N.Y.

William E. Sollecito, '53; November 2, 1984; Syra

cuse, N.Y.

John B. Padgett, Jr., '54; July 16.1984; Palos Verdes,

Calif.

David R. Wones, '54; December 1984; Blacksburg,

Va.

William B. Banks, 55; February 23, 1985; Port Or

ange, Fla.

Thomas C. Wood, '55; April 19, 1984; Manhattan

Beach, Calif.

Robert R. Pollard, '56; January 29, 1985; Hollis,

N.H.

Harry M. Salesky, '57; 1985; Tiburon, Calif.

Robert Van Benschoten, '57; March 15, 1984; Liv

ingston. N.J.

Donald Daryl Wyckoff, '58; January 20, 1985; Mar-

blehead. Mass.

Bruce R. Hayworth, '59; January 20, 1985; Poway,

Calif.

Maciej James Achmatowicz, '64; October 28. 1984;

Ashburn, Canada.

Ralph W. McKenney, Jr., '66; January 18, 1985.

Stanley D.Derbin, '69; November 29, 1984; Dan-

bury, Conn.

Kenneth R. Britling, 71; February 18, 1985; Dover,

Mass.

Omer S. Kaymakcalan, '75; February 22, 1985; De-

wilt, N.Y.

David A. Anderson, 76; February 22, 1985; Ha-

worth, N.J.

Wayne E. Matson, '77; February 1985; Boston.

Randall G. Chipperfield, '85; Cambridge, Mass.

Karl N. Horita, '85; January 25. 1985, Brookline.

Mass.

AlexanderKusko, lite

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Speed Department

SD 1. Phelps Meaker is building astraight sidewalk 54'8" long, using 40

cast-concrete slabs each formed as an

equilateral triangle. To square off the

ends, he has two extra 30-60-90 half-slabs. What is the width of the walk?

SD 2. A bridge quickie from Doug VanPatter:

Your

A

¥

♦

*

A

A

K

A

hand:

K10

QJ1072

Jio

Dummy:

AJ53

V 965

♦ AJ74

* 062

Your contract is six hearts. West leads alow club. You put in the #Q, which is

covered by the *K and *A. Can you

find a line of play that just about guarantees success?

Solutions

FEB/MAH 1. Management meetings are scheduled

on the second Thursday of each month, adminis

trative conferences are the third Friday, and work

units have a seminar on the first Monday. Derive

an alogrithm which will generate a date given the

year, month, day-of-week, and ordinal week within

the month. For example, if a meeting were scheduled for the third Friday of August 1984, the algo

rithm would return "August 17." Note that a

meeting on the fifth Tuesday in March would be

fine for 1983, 1985, and 1986 but not for 1984 (thereare only four Tuesdays in March of 1984). In this

case the algorithm could return January 0.

The following solution is from Al Weiss, who also

notes that there are only four Tuesdays in March

in 1985 and 1986:

Programmers, like all good craftsmen, have their

own bags of tools. One tool that has been in my

bag for a long time is a routine that will convert a

date into the number of days since November 24,

-4713 (this is not the same as 4713 B.C., since there

was no year zero). The number generated by this

routine is called the Julian day. It is similar to a

"shop date," which is the number of days since the

beginning of the year. This number itself is not too

useful, but it can be used to calculate the number

of days between two dates. There is also a com

panion routine which converts the number back

into a date. These two routines can be used of find

the date 90 days from today. Using these two rou

tines it is possible to solve Alfred Anderson's problem. My procedure is as follows:

1. First we determine the Julian Day of the 1st of

the month desired (assuming we are looking for the

third Friday in July 1985, this would return the number 2446248).

2. We then determine what day of the week this is

(in this case the program returns a 1 telling us that

this is a Monday). Next we find the date of the firstFriday (the computer tells us that this is the 5th).

3. Then we calculate the date of the 3rd Friday (inthis case the 19th).

4. Finally the program validates the answer by cal

culating the Julian day of this dale and then reconverting the Julian day back to a real date. If themonth of this new date is the same as the original

month, we have a solution. As an example of an

impossible date: if we were looking for the 5th Fri

day in July, the computer would calculate.it to bethe 33nd of July. When this is converted to a Julian

day and then back to a real date it comes back as

August 2nd. The computer recognizes that August

is not the month requested and sets the date to zero.Mr. Weiss submitted the program described

above, but inadequate space prevents us from reprinting the program and we apologize. A copy can

be had by return mail from the Review.

Also solved by Frank Carbin, Harry Zaremba, JimLandau, John Patterson, Matthew Fountain, Robert

Slater, Winslow Hartford, and James Abbott (whoalso included a Tl-59 program card containing hissolution.)

FEB/MAR 2. Two coins, loosely coupled, are flipped

simultaneously such that if either one is heads, theother has probability 7/8 of also being heads, but ifeither one is tails, the other is equally likely to be

either heads or tails. Find the probability of eachindividual coin turning up heads, and the probability of their both being heads simultaneously (or

prove that the problem statement and data are inconsistent).

Michael Tamada found a couple of solution techniques for the loosely coupled coins:

Each coin has probability of .8 of being heads. Theprobability that both are heads is .7:

Coin I

I have two ways of deriving the answer, one usinga "contingency table" approach and one using a"conditional probability" approach.

Contingency Table Approach:

We wish to find the unknown probabilities a, b, c,and d:

H

T

Co

H

.7

.1

.a

in 2T

,|

.1

.2

.8

.2

Coin 2

Coin I "-

H

o

c

T

b

d

We know that a + b + c + d = l. If Coin 1 is tails(i.e., if we're in the second row, which contains c

and d), we are told that Coin 2 has equal probabilities of being heads or tails. In other words, c/(c +

d) = 1/2. Similarly, if Coin 2 is tails, we are told

that b/(b + d) = 1/2. These two equations tell usthat b = c = d, so we know that a + 3b = 1. We.are told that if Coin 1 is heads, then Coin 2 has a

probability of 7/8 of being heads. In other words,

a/(a + b) = 7/8. Substituting a = 1 - 3b in theabove equation, we get (1 - 3b)/(l - 3b + b) =

7/8, or b = .1. Since b = c = d, we know c = d

= .1 and thus a =.7.

Conditional Probability Approach:

Let "HI" and "Tl" stand for the probability that

Coin 1 is heads or tails respectively. Obviously HI

= 1 - Tl (and H2 = 1 - T2 for Coin 2). Letp(Hl:H2) stand for the probability of event HI given

that H2 occurs. We are told that p(Hl:H2) = 7/8

and p(H2:Hl) = 7/8. Similarly, we are told that

p(Hl:T2) = 1/2 and p(H2:Tl) = 1/2. The uncondi

tional probability of any event is equal to the sum

of its conditional probabilities (weighted by the

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probabilities of the condition occuring). I.e.,

p(Hl) = p(Hl:H2) x p(H2) + p(Hl:T2) x p(T2).

So

p(Hl) = (7/8)p(H2) + (1/2)|1 - p(H2)]

p(Hl) = 1/2 + <3/8)p(H2). (1)

Similarly, for Coin 2 we get

p(H2) = p(H2:Hl) x p{Hl) + p(H2:Tl) x p(Tl)

p(H2) = <7/8)p(H) + (1/2)[1 - p(Hl)J

p(H2) = 1/2 + (3/8)p(Hl). (2)

Combining equations (1) and (2),

p(Hl) = 1/2 + (3/8)[l/2 + (3/8)p(Hl)] = 4/5.

We also find that p(H2) = 4/5. Again using laws of

conditional probability, we know that

p(Hl & H2) = p(Hl:H2) x p(H2) = 7/8 x 4/5 =

Also solved by David DeLeeuw, Leon Tabak,

Matthew Fountain, Michael Jung, Richard Hess,

Winslow Hartford, and the proposer, William Stein.

FEB/MAR 3. A horizontal line of length 2a forms

the common base for two isosceles triangles. The

near side the triangle is 45° -45" -90°, and on the

opposite side 75° -75° -30°. Determine the radius

of the circle tangent to all sides of the composite

lanceolate figure, and locate the center.

Avi Omstein makes it look easy:

Let x be the radius of the inscribed circle, let h be

the line segment from the circle's center to the ver

tex of the 75° -75° -30° triangle, and let s be the

length of the bisector of this triangle. From the dia

gram, we see:

s = a/tan 15° and h - x/sin 15"

a + s = x21/2 + h.Thus we have

a + a/tan 15° = x21/2 + x/sin 15°

a(l + 1/tan 15^(2"* + 1/sin 15°)x = 0.896575472a

x2w = 1.267949192a

Also solved by David DeLeeuw, Everett Leroy,

George Parks, Harry Zaremba, Mary Lindenberg,

Matthew Fountain, Mel Garelick, Naomi Markov-

itz, Richard Hess, Steve Feldman, Winslow Hart

ford, and the proposor, Phelps Meaker.

FEB/MAR 4. Find a four-digit number whose square

is an eight-digit number whose middle four digits

are zero.

Most solutions were brute-force computer

searches. Pierre Hefller reduced the search vastly

by employing some pre-analysis:

The answer is 6,245, whose square is 39,000,025.

Trivial answers of 4,000, 5,000, 6,000, 7,000, 8,000

and 9,000 should have been excluded in the state

ment of the problem. Since the square must lie be

tween 10,000,001 and 99,000,099, the number itself

must lie between 3,163 and 10,000. Absent any way

to predict the occurrence of zeros in the middle of

a square, one could square each number between

3,163 and 10,000 (6,144 numbers in ail if endings in

zero are omitted) and hope to find a square with

four zeros in the middle. A tedious search without

a computer, almost two hours on my HP97. For a

more efficient search, consider the following: the

square root of 10,000,099 is 3,162.293; the square

root of 10,000,000 is 3,162.277. The difference is

0.0156. In the same test for numbers 99,000,099 and

99,000,000, the difference in square roots is 0.00498.

It follows that if X is a number ranging up from

10,000,000 and if VX comes out with a fractionwhich is more than 0.0156, there is no number be

tween X and X — 99 which is a perfect square. So,

using 0.0156 as a discriminant, take the square root

of 90 numbers of the form ab,0O0,O99, where ab

ranges from 10 to 99 (easy on any pocket calculator),

discard any square root if its fractional part exceeds

0.0156, and round out the rest to the next lowest

whole number. Then discard any left that end in

zero. Three remain that are worthy of being tested.

One of them, 6,245, satisfies the problem. The other

two do not because the discriminant did not de

crease to 0.00498 (as it should have to be a necessary

and sufficient test) as ab ranged up to 99. This search

took just over two minutes on my HP97.

Also solved by Jerry Cogan, Frank Carbin, Ches

ter Claff, Avi Omstein, Dennis Sandow, George

Byrd, George Parks, John Prussing, Lee Fox, Mat

thew Fountain, Michael Jung, Michael Tamada, Na

omi Markovitz, Nicholas Strauss, Richard Hess,

Robert Slater, Robert Turner, Ronald Raines,

Thomas Stowe, and Winslow Hartford.

FEB/MAR 5. Consider two dipoles. The lower di-

pole is fixed, and the upper dipole is constrained

to move along a horizontal line. (This is roughly the

geometry encountered in magnetic stirring.) Find

the conditions for which the upper dipole tends to

center (the force is in the opposite direction to the

displacement from the center line). When does the

motion of the upper dipole approximate simple har

monic motion?

Only Matthew Fountain and Richard Hess tac

kled this hard problem. Mr. Fountain's impressive

solution would have been published except for a

confrontation between its length and the available

space in this issue. Readers may obtain a copy (to

gether with our apologies, which go also to reader

Fountain) by return mail on request to the Review

office.


JAN 4. Pierre Heftier has responded.

Proposers' Solutions lo Speed Problems

SD 1. 27.733

SD 2. First cash the *K, then lead the VQ. If this

holds (it did) and both defenders follow suit, play

the */J. Presumably, a defender will take the VK.

Now the ¥9 provides an entry to dummy, and you

can pitch the A10 on the ♦A. If East shows oul on

the first trump trick, play a small trump toward the

V9, with the same result. This hand occurred at

the Talleyville Club in Wilmington, Del., which

boasts of the highest percentage of life masters in

the U.S. West actually held three hearts to the king.

Several declarers made 12 tricks.

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PUZZLE CORNER

ALLAN J. GOTTLIEB

Fun with Trig

Since it has been over a year since I

reviewed the criteria used to select

solutions for publication, let me do

so now.

As responses to problems arrive, they

are simply put together in neat piles,

with no regard to their date of arrival or

postmark. When it is time for me to

write the column in which solutions are

to appear, I first weed out erroneous and

illegible responses. For difficult prob

lems, this may be enough; the most pub-

lishable solution becomes obvious.

Usually, however, many responses still

remain. I next try to select a solution that

supplies an appropriate amount of detail

and that includes a minimal number of

characters that are hard to set in type.

A particuarly elegant solution is, of

course, preferred. I favor contributions

from correspondents whose solutions

have not previously appeared, as well

as solutions that are neatly written or

typed, since the latter produce fewer

typesetting errors.

Finally, let me credit David Griese-

dieck as the solver of OCT 3. Somehow

his name disappeared when the column

travelled from New York to Cambridge.

Problems

M/Jl. We begiii with a two-part bridge

problem from Doug Van Patter based on

tho following deal:

A K

VQJ

♦ Q9

+ Q

♦

♦

*

9432

832

A

¥

*

A84

AK10

A765

KJ

9765

7

K10

A986

4

♦ QJ¥86

♦ J+ 10/

32

10 3 2

5

'54

How does South make five clubs after

an opening lead of the ♦ Q, and what

lead sets this contract?

M/J 2. As I look out my kitchen window

and see our lake frozen solid, I am

warmed by Phelps Meaker's N-sided

flower garden described in the following

problem, entitled "Fun with Trig":

A large bed of flowers and greenery

is laid out in the form of a regular pol

ygon of N sides. A walk composed of N

trapezoidal concrete slabs surrounds the

flower bed. A circumscribing circle pass

ing through the outer corners of the

walk and an inscribing circle tangent to

the inner flats of the trapezoids have cir

cumferences in the ratio of almost ex

actly 191:165. The total area within the

outer periphery of the walk and the area

of the walk itself are in the ratio of 4:1.

Find N.

M/J 3. Lester Steffens wants you to ex

plain a parlor trick he is fond of:

Given a number N between 500 and

1000, rapidly construct a series of num

bers using eight of the nine digits from

1 through 9 once, the other digit twice,

and a few zeros so that the series totals

N. Example: if N = 642, a solution is

10

20

334

50

60

78

_90

642

M/J 4. Winslow Hartford asks a mniii-

faceted problem:

As we generate geometric figures to

represent y = x", we have "elements"

consisting of points, lines, faces, cubes,

etc. as n increases. For the number of

points in each figure, we have (for n >

0) P = 2". Derive the number of lines

and faces in a five-dimensional hyper-

cube.

M/J 5. Eric Schonblom tells us about his

8.01 doodles, with an apology for tar

diness: "Having discovered this during


AND COMMENTS TO ALLAN I.

GOTTLIEB. -67. ASSOCIATE




NEW YORK UNIVERSnr. 251


N.Y.. 10012.

A22 MAY/JUNE 198.5

OBITUARIES

Highway Safety Pioneer

Lived a Life "Deserving of Prominence

A TRIBUTE BY RALPH NADER

William Haddon Jr.

1926-1985

Physician William Haddon, Jr., '49, a leader

in highway accident research, died in Wash

ington, D.C. on March 4 of the complica

tions of kidney failure. Consumer advocate

Ralph Nader believes that Dr. Haddon's ca

reer owed much to his M.l.T. education and

that his death at 58 was a tremendous loss

to the cause of highway safety. In many

ways, Dr. Haddon's life exemplified the val

ues propounded by Nader when he gave the

keynote address at the Alternate Jobs Fair,

organized in late winter by the M.l.T. Stu

dent Pugwash (see facing page).

William Haddon, Jr.

"T oss prevention" is not a conven-

I tional focus for engineers and phy-J—Jsicians, but Dr. William Haddonperformed so brilliantly in that field that

students aspiring to technical and med

ical careers can benefit greatly from an

awareness of his contributions.

After obtaining degrees from M.l.T.,

Harvard Medical School, and the Har

vard School of Public Health, Haddon

spent 10 years with the New York State

Department of Public Health. He was

concerned with highway traffic casual

ties, which at that time attracted few

minds possessing staying analytic

power and follow-through stamina,

traits Haddon had in ample supply.

Motor vehicle crashes and trauma

were failures in energy management, in

Haddon's view. He separated the ve

hicular crash phenomenon into pre-

crash, crash, and post-crash stages in

order to clarify the intervention appro

priate for each stage. Haddon demon

strated how the engineering

intervention known as crashworthi-

ness—exemplified by air bags—could

manage the lethal transfers of energy in

a crash to save life and limb. Through

his studies on alcohol-related crashes,

he was familiar with those data. But he

believed that the engineering approach

had a greater chance of reducing death

and injury than any available method for

controlling the complex behavioral

sources of drinking and driving.

This emphasis on engineering reme

dies was rooted in Haddon's knowledge

of empirical successes throughout his

tory (which he cited frequently)—from

the Greek physician Hippocrates to ad

vances in factory and railroad safety for

workers—and in his detailed knowledge

of crash protection systems which were

available but not in use.

In the 1960s, Haddon was part of the

growing consumer challenge to the au

tomobile industry though his writings,

testimony, and television and radio in

terviews. From 1966-1969, he was the

first administrator of the National High

way Traffic Safety Agency established

by President Johnson. Under Haddon's

direction, the agency issued motor ve

hicle safety standards, conducted re

search, and required the recall of

defective vehicles and tires.

After leaving government, Haddon

became the president of the Insurance

Institute for Highway Safety in Wash

ington, where he took the lead in trans

forming the insurance industry into a

critic of unsafe and costly vehicular en

gineering. He also helped establish the

Highway Loss Data Institute, which

provided safety information on vehicles

by make and model.

The words that come to mind when

reflecting on Haddon's work are rigor

ous, dedicated, unyielding in purpose,

humane, blunt, and a long-distance run

ner. He traversed the whole contin

uum—making major contributions to a

conceptual foundation for trauma pre

vention, to public education, safety reg

ulation and research, and to a new

perception of the mission of the insur

ance industry. Haddon also raised the

status of professional work in trauma

prevention—introducing younger mem

bers of relevant professions such as sta

tistics, engineering, and biology to the

cause of saving lives through knowledge

and its application.

Haddon's was a life to be celebrated,

deserving of prominence, in the hope

that many will follow in his footsteps if

they only know of his trail. □

DeceasedThe following deaths have been reported to theAlumni Association since the Review's last deadline:

John B. Welch, '13; October 3, 1984; Indianapolis,Ind.

Atwood P. Dunham, '17; December 14, 1984; Ded-ham. Mass.

Raymond E. McDonald, '17; October 18, 1984; Na-lick. Mass.

Freeman H. Dyke, '20; March 8, 1984; Tequesta,Fla.

Fraser M. Moffat, '20; October 24, 1984; Montrose,Penn.

William D. Shepard, '20; January 1985; Winnetka,

James J. Wolfson, '20; November 14, 1984; Hallan-

dale, Fla.

Chester A. Rimmer, '21; February 7, 1985; Norwell,Mass.

George T. Boli, '22; 1985; Venice, Fla.

John F. Hennessy, '22; February 9, 1985; ChestnutHill, Mass.

W. Raymond Hewes, '22; December 24,1984; Need-ham, Mass.

James D. Sarros, '22; December 27, 1984; Madison,N.J.

Mrs. Francis W. Spalding, '22; July 1984; Cincinnati, Ohio.

Arthur R. Belyea, '23; December 16, 1984; Old Say-

brook, Conn.

Arthur Raymond Holden, '23; February 11, 1985;

Sarasota, Fla.

PHOTO: INSURANCE INSTITUTE FOR HIGHWAY SAFETY TECHNOLOGY REVIEW A21

a physics lecture over 30 years ago, I'm

a little slow in sharing it. It's a paper-

and-pencil puzzle but is most easily

stated in terms of scrabble tiles":

Take the letters in the first half of the

alphabet in order. Place the A on the

table. Place the B next to one of the four

sides of the A. Place the C next to the

one of the six sides at the AB (or BA)

pair. Then add the D and so on. If you

do this correctly, when you reach the

letter M you will have created a cross

word-puzzle matrix of complete com

mon English words, no proper names,

no foreign words, and no acronyms or

abbreviations. Having solved the prob

lem as posed, can you add one or more

letters to the A-M set and still retain

complete words?

Speed Department

SD 1. Jerry Grossman wants to know

what is so interesting about

f(x) = x1***,

where log is the natural logarithm.

SD 2. David Evans designed a single

elimination tennis tournament for 37

contestants to have the minimum num

ber of byes. How many matches were

played?

Solutions

JAN 1. South is on lead and is to take six tricks

against best defense with hearts as trump:

A 9832

V 10

♦ AQ

* —

J¥ 8

♦ ) 10 9 8

♦ 5

* 10

V 6

♦ K6

AKJ6*AK7y

♦ —

* AQ4 3

Ben Feinswog solved the problem by losing the

A7 and discarding the * A and *K:

South plays the *A (removing West's exit), dis

carding the ♦ Q from dummy, and then leads the

*7 to West's A). On the forced red-card return.

South plays dummy's winning ¥10, and ♦ A, dis

carding the *A and 4>K from hand, and claims the

balance with dummy's last three spades.

Also solved by Avi Ornslcin, Doug Van Patter,

Edgar Rose, Ellen Kranzer, Joe Hahn, John Lacy,

Larry Wischhoefer, Matthew Fountain, Red Clev-

enger, Robert Lax, Roy Schweiker, Tim Maloney,

Walter Cluett, Winslow Hartford, Jim Landau, John

Rule, Richard Hess and Emmet J. Duffy.

JAN Z. A smooth, rigid, and circular hoop hangs

from a rigid support by an ideal, extensionless

string. Two small beads slide along the hoop (like

beads of a necklace) with negligible drag and fric

tion. The beads are slid to the top of the hoop and

released. How massive must each bead by lo spon

taneously lift the hoop?

Matthew Fountain was pleased to submit a so

lution but even more pleased to report that his wife

has responded perfectly to a recent cataract oper

ation:

The two beads must total three rimes the mass of

the hoop. As each bead slides outward, its circular

motion causes a centrifugal force with an upward

component. At the same time its downward accel

eration increases, decreasing the force that it exerts

upon the hoop. The maximum lifting force occurs

when the sum of these two effects is greatest. Until

the hoop actually moves, a bead does no work.

Therefore, its gain in kinetic energy equals its loss

in potential energy. Thus, (l/2)mvy =■ smg, wheres = vertical drop, m - mass, g - gravitational

constant, and v = velocity.

\When the bead has traversed an arc <(> on the hoop

of radius r, the vertical drop

s = r(l - cos *).

The centrifugal force is mv*/r, with an upward component

(mvVr)cos 4> = 2smg(cos <l>)/r = 2mg(l -cos <|>) cos

<t> . Gravity produces a force mg acting downward

through the bead. When the bead has traversed

through the arc 4>, the component of this force

toward the center of the hoop is mg(cos<t>). In rum,

the downward component of the radial component

is mg(cos! <)>). The lifting force F exerted by a beadis

F = 2mg(l — cos <t>)cos <J> — mg(cosJ 6) = 2mg(cos

<f>) - 3mg(cos2 *).

The maximum and minimum values of F occur

when

dF/d<J> = - 2mg(sin 4>) + 6mg(cos <J>)(sin 4) = 0.

Thus the minimum lift occurs when sin 6 = 0 and

the maximum lift occurs when cos <S> = 1/3:

F™, = (l/3)2mg - (1/3)2 3mg = (l/3)mg.

Each bead will lift up to one-third of its own weight.

Also solved by David Smith, Gary Heiligman,

Harry Garber, Harry Zaremba, John Lacy, John

Prussing, Ken Haruta, Jim Landau, Richard Hess,

Peter Kramer, and the proposer, Bruce Calder.

JAN 3. A man received a check calling for a certain

amount of money in dollars and cents. When he

went to cash the check, the teller made a mistake

and paid him the amount which was written in

cents in dollars, and vice-versa. Later, after spend

ing S3.50, the man suddenly realized that he had

twice the amount of money the checked called for.

What was the amount on the check.

Red Clevenger provides a "down home" solu

tion:

The problem reminds me of my eighth-grade

teacher in rural Afton, Okla., who had several dol-

lar-and-cents algebraic problems which she always

wanted us to solve by using one variable. I preferred

using two variables in simultaneous equations. She

did drive home the lesson, however, of lOOtf = 15.

Let

$ = the dollar amount of the check, and

c = the cents amount;

then the problem stated algebraically is:

lOOe + $ - 350 = 2(100$ + c). (1)

The other equation results from the difference be

tween the number of cents received (S) and twice

the number of cents on the check (c) which must

be - 50 since c is greater than S. Staled algebraically:

S - 2c = - 50. (2)

Solving (1) and (2) results in S = 14 and c - 32,

and thus the check amount was SI4.32.

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Qtorglo Pfccao.II,

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San Francisco, CA94112

(415)333-5084

Gecnge A.Roman &Associates Inc


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Horl P. Aldrich, Jr. '47

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Douglas G. Grrford 71

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John P. Ougan '68

Kenneth L Rocker '73

Mark X. Haley '75

Robin B. Dill '77

Andrew F. McKown 78

Keith E. Johnson 'SO

Also solved by Winslow Hartford, Allan Benson,

Anthony Lombardo, Avi Ornstcin, Charles Sutlon,

David Smith, Edgar Rose, Frank Carbin, Fred Sleig-

man, Gary Drlik, George Byrd, George Aronson,

Harry Garbcr, Harry Zarcmba, Howard Stern,

James Michelman, John Lacy, John Prussing, Ken

Haruta, Larry Wischhoefer, Leon Tabak, Marion

Berger, Matthew Fountain, Michael Jung, Naomi

Markovitz, Norman Spencer, Peter Silverberg,

Phelps Meaker, Ronald Martin, Steve Feldman, Ted

Numata, Thomas Stowe, Everett Leroy, Tim Ma-

loney, Jim Landau, Ronald Raines. Richard I less,

George Parks, and the proposer John Rule.

JAN 4. A rigid arm pivots around the fixed point

A. At the end of the arm is a follower (B) which

runs in a curved track. The track pivots about the

fixed point C. If AB = AC = r, find the shape of

the track such that its slope at C is always vertical.

The problem as stated was rather easy. Surpris

ingly, several readers noticed that the diagram in

dicated a vertical slope at B, not C. This was indeed

more interesting. Charles Sutton writes:

The problem as stated is trivial, since clearly a track

in the shape of a semi-circle of radius r with center

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at A would remain stationary and its slope at C

would always be vertical. I assume what was in

tended was that the slope of the track at B should

always be vertical. This really had me going for a

while. I set it up in rectangular coordinates, using

anlaytic geometry and calculus, and ended up with

an impossible differential equation. Then [ tried po

lar coordinates, used a few trig identities and cal

culus, and got a real simple differential equation

that integrated to give me a circle. Once you know

what you're looking for, you need only elementary

geometry. For the slope at B to remain vertical, the

shape of the curved track is determined by the facts

that AC = AB = r and the tangent to the track at

B is perpendicular to the line at AC. Imagine that

the track is held fixed; then point A will have to

move on a circle of radius r. The accompanying

diagram shows that the track is an arc of a circle of

radius r with center at D. The tangent to the circle

at B is perpendicular to both BD and AC, so AC

must be parallel to BD. Also AC = BD - r, so ABCD

is a parallelogram and AB = CD - r.

Also solved by David Smith, Gary Heiligman,

George Byrd, Harry Garber, Harry Zaremba. How

ard Stem, John Lacy, Jordan Wouk, Matthew Foun

tain, Phelps Meaker, Red Clevenger, Winslow Hart

ford, Richard Hess, Peter Kramer, and the pro

poser, Floyd Kosch.


Y19M Randall Whitman, William Thompson, Rick

Lufkin, Marion Berger, Al Weiss, Matthew Foun

tain, Donald Trumpler, and Alan Katzenstein each

improved on the published solution. When com

bined, their efforts yield the following list of revi

sions:

26 = (1 + 9/4) x 8

31 = 49 - 18

37 = 1 + 9(8 - 4)

41 = 49 - 8 x 1

42 = (49 - 8) + 1

49 x Is

81 - (4 + 9)

V x 8/4

4-1"

1" x 4

1" + 4

(49 - l)/8

91-84

8 x 1«

= 1" + 914 = 14 x (9 - 8)

16 = (1 + 9 - 8)4

18 = 9 x 4 - 1822 = (89 - l)/4

49 =

68 =

70 =

76 =

81 =

99 =

100

(9 - 1/4) x 8

94-18

4 x 18 + 9

98 + I4

= (1 + 9**

JUL 2. Howard Stern notes that this problem ap

pears in Martin Gardner's Mathematical Circus.

OCT 3. John Stackpole notes that the very week his

F/M 1 issue of Technology Review arrived, the Mary

land Lotto game had ten participants match all six

numbers and 827 match five out of six, giving

strength to Griesedieck's assertion that the selection

of numbers by participants is not random. Jonathan

Hardis, the proposer, remarks that never have any

of the large lotteries had a positive expected return

for the player and also expresses the oft-stated po

litical belief that these games constitute a regressive

tax since they played to a disproportionately large

extent by the poor.


SD 1. It is constant, i.e. independent of x.

SD 2. 36 One contestant is eliminated each match.

A24 MAY/JUNE 1985

PUZZLE CORNER

ALLAN J. GOTTLIEB

Can You Find the Bad Penny in the Bank?

Just as I am writing this the mailman has

brought a letter from the New York State

Crime Laboratory marked "official busi

ness." After two gulps and one re

flection as to what I could have done that

they found out about, I got up my cour

age to open the envelope and was re

lieved to find it was just a "Puzzle

Corner" reader responding to several

problems from the January issue.

As I have remarked previously, Nob-

uyuki Yoshigahara selected me as

"World Puzzlist" No. 8, and now he has

forwarded the issue of Quark in which

this honor was officially bestowed. In

addition to being flattered, I enjoyed

hearing the transliteration of "Gottlieb"

when one of my Japanese-speaking col

leagues read the beginning of the col

umn. Thank you again, Mr.

Yoshigahara. I should also mention that

long-time "Puzzle Corner" contributor,

Richard Hess, was selected World

Puzzlist No. 6.

Problems

APR 1. I read our first problem for this

month in net.chess, an electronic news

group devoted to chess. Roughly speak

ing, these newsgroups consist of widely

separated individuals who communi

cate with each other via electronic mail.

I especially enjoyed the following two-

part offering from Jeffrey Mattox, who

noted that it is possible, albeit unlikely,

for the position to occur in a game:

White is to play and mate in two. Mattox

notes that at first glance there appears

to be two possible solutions. You are to

show that only one meets the need.

APR 2. Our next problem is from Phelps

Meaker, who first asks you to study:

16601.92 + 14374.08 =

11334.4 + 19641.6 =

18521.44 + 12454.56 =

4147.36 + 26828.64 =

He then notes that each pair is a differ

ent way of evaluating the same equation

and asks you to write the equation in

the usual form. He also offers a hint, but

you may wish to try the problem with

out this aid. [The hint is to note which three

numbers are perfect squares.]

APR 3. Rich Decker wants you to find

all maxima and minima of

without using any calculus. This problem

appeared in an Ohio State University

prize exam for undergraduates.

APR 4. Allan Faller wants us to be penny

wise and writes:

On each day of the year (not leap year)

you are given a penny. On December 31

you are given your last penny and told

that it was fresh from the U.S. Mint, but

that one of the previous pennies may

have been counterfeit, and therefore

lighter or heavier. than the standard

penny. You are asked to determine the

number of balancings, using a common

pan balance, that would be necessary

and sufficient to determine whether or

not there is a counterfeit coin, and if

there is, to tell whether it is heavier oflighter than the last penny that you re

ceived.

APR 5. Our final regular problem, from

Martin Brock, is based on the familiar

"crossed ladders" configuration at the

bottom of the previous column:

Mr. Block asks you to find X for two

configurations. First when A = 15, B =

10, and H = 8; and second when A =

16 + 2V2, B = 16 - 2V2, and H = 2.

Speed Department

SD I. A bridge quickiePaftor*1 atier.

North:

A K854

¥ Q 10 7 4

♦ 53

♦ AQ 10

South:

♦ A93

¥—

♦ AQ6

*K987643

East: South:

IS 2C

3H 5C

5H 6C

West

P

P

D

from Doug Van

: North:

3C

P

P

Instead of defending the usual five-heart

bid by East, you (South) make the ag

gressive bid of six clubs. West opens

with ¥5, which draws the ¥10, ¥K,

and a trump. You lead a club to dum

my's «fcA and East shows out. Your

finesse of the *Q loses to West's 4K,

and West returns a trump to the *10.

You lead to the 4A (East shows out),

and ruff your third diamond with dum

my's last club. Can you find a way to

justify your overbid? (East is an excellent

player, never known to psych).

SD 2. Joseph Horton writes: Great news!

I have answered an age-old question:

Which came first—chicken or egg?

SEND PROBLEMS, SOLUTIONS.


GOTTLIEB. 67. ASSOCIATE




NEW YORK UNIVERSITY. 251


N.Y.. 10012.

A.)0 APRIL 1985

■ OBITUARIES,

Jerome H. Holland,

1916-1985

Jerome H. Holland, prominent black

educator and civil rights advocate

who was a member of the Corpora

tion from 1969 to 1979, died in New

York on January 13 of cancer; he was 69.

Dr. Holland studied sociology at Cor

nell and the University of Pennsylvania

and went on to become president of two

predominently black colleges, Delaware

State College and later Hampton Insti

tute, before accepting appointment as

Ambassador to Sweden.

In addition to his service on the M.I.T.

Corporation and its visiting committees

for the Center for International Studies,

Department of Humanities, and ath

letics and student affairs. Dr. Holland

was national chairman of the American

Red Cross and the first black on the

board of the New York Stock Exchange.

Deceased

Robert S. Beard, 05; 1983; Trinidad, Calif.

Mrs. Milton E. Hayman, 11; 1984; West Hart

ford, Conn.

Clifford L. Muzzey, Sr., '14; December 15, 1984;

Sandusky, Ohio.

Solomon Schneider, '15; October 12, 1984; Haver-

town. Penn.

James B. Hobbs, '16; January 6, 1985;. Natick,

Mass.

Donald B. Webster, 16; November 25, 1984; Fal-

mouth, Mass.

Benjamin I. Lewis, '17; December 21, 1984; Ta-

coma, Wash.

R. Parry Kennard, '18; 1984; New York, N.Y.

John S. Coldwcll, 19; September 21, 1984; Fall

River, Mass.

Shee M. Lee, 19; 1984; Hsinchu, Taiwan.

Chester C. Stewart, 19; May 18, 1984; Needham.

Mass.

George H. Wiswall, Jr., 19; 1982; Edgartown.

Mass.

Preston W. Smith, '21; January 6, 1985; North

Weymouth, Mass.

Charles Kerr, Jr., '22; Fort Point, Fla

Dwight E. Stagg, '22; November 13, 1984; Bridge

port, Conn.

William P. Winsor, '23; December 16, 1984; New

York, N.Y.

Cordon H. Crabb, '24; August 1, 1984; Winter

Park, Fla.

Mrs. Theodore G. Coyle, '25; July 12, 1984; Nor

wood, Mass.

Marvin H. Green, '25; November 6, 1984; N.

Palm Beach, Fla.

Eugene C. Hermann, '25; August 20, 1984; West-

field, N.J.

George Oetinger, Jr., '25; 1984; Monticello, Fla.

Peter H. Sin, '25; September 13, 1984; Hong

Kong.

Mrs. George V. Slottman, '25; January 5, 1984;

New York, N.Y.

Elmer C. Warren, '26; October 11, 1984, Waler-

ville, Maine.

Adelbert N. Billings, '27; November 13, 1984;

Richmond, Va.

Laurence H. Coffin, '27; November 11, 1984;

North Conway, N.H.

George D. Fexy, '27; September 1984; Kirkland,

Wash.

Randolph J. Peterson, '27; November 4, 1984;

Rochester, N.Y.

John M. Ryan, '28; November 3, 1984; Winches

ter, Mass.

John D. McCaskey, '29; January 10, 1985; St. Jo

seph, Mo.

Theodore Criley, Jr., '30; January 1985; Clare-

monl, Calif.

Norman H. Dolloff, '30; October 3, 1984.

Harry J. Fekas, '30; November 2, 1984; Newport

News, Va.

Emile P. Grenier, '31; December 12, 1984; Ann

Arbor, Mich.

John Vasla, '31; 1984; Punta Gorda, Fla.

George E. Colby, '32; October 3, 1984; Westport

Harbor, Mass.

John F. Crowther, '32; October 19, 1984; Old

Lyme, Conn.

George Hcnning, '33; November 21, 1984; Syos-

set, N.Y.

Walter S. Brodie, '34; January 20, 1985; Marble-

head, Mass.

Karl A. Gardner, '34; January 3, 1985; North-

ridge, Calif.

Herbert L. Gamer, '34; November 26, 1984; Mil-

Ion. Mass.

Proctor Wetherill, '34; January 16, 1985; Chester

Springs, Penn.

Stephen H. Richardson, '36; 1984; Seattle, Wash.

Robert M. Sherman, Jr., '36; November 27, 1984;

Warwick, R.l.

Staunton L. Brown, '38; October 8, 1984; Madi

son, Conn.

Roscoe J. Cooper, '38; August 27, 1984; Beverly,

Mass.

Edgar H. Kiblcr, Jr., '39; 1983.

Michael Morelli, '40; November 16, 1984; Alexan

dria, Va.

George Farneil, '41, September 2, 1984; Syracuse,

N.Y.

J. Nelson Evoy, Jr., '42; November 1984; Bryn

Mawr, Penn.

Howard W. Comey, '43; 1984; South Acton,

Mass.

Howard Weaver, Jr., '44; 1983; Phoenix, Ariz.

Edmond J. McClure, Jr., '45; 1983; Blasdell, NY

James W. Shearer, '45; October 9, 1984; Liver-

more, Calif.

Charles H. Tavener, Jr., '45; Jauary 17, 1982, Boca

Raton, Fla.

James W. Church, '46; 1983; Silver Spring, Md.

Norman P. Hobbs, '48; October 17, 1984; Chest

nut Hill, Mass.

George H. Bradley, Jr., '49; April 19, 1984; Albu

querque, N.M.

Wilmer S. Garwood, Jr., '50; December 8, 19S4

Ben Silver, '50; 1982, Barrington, R.I.

James B. Law, '52; 1984; New Delhi, India.

Vernon V. Hukee, '56; September 26, 1984; Na

shua, N.H.

William L. White, '59; January 21, 1985; Newton-

ville. Mass.

Mohammad R. Damghany, '62; December 22,

1984; Arlington, Mass.

William C. Margetts, '62; 1984; Cambridge, Mass.

Edward Terner, '65; September 27, 1984; Shel-

bume, Vl.

Louis J. Urban, '70; October 31, 1984; Kettering,

Ohio.

James S. Steil, '80; December 20, 1984; Moses

Lake. Wash.

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Solutions

N/D 1. While, moving first, is to mate in threemoves:

Howard Stem found this problem to his liking:

The following three plays represent the only possibilities resulting from While's initial move:

While:

Q-C2Kn-d3

Q-b2 mate

While:

Q-c2

Kn-f3

Q-d2 mate

While:

Q-C2

Kn-d3

Q-a2 mate

Black:

KM

K-a3

Black:

K-dA

K-e3

mack:

K-b4

Also solved by Eric Rayboy, David DeLeeuw,

R. Bart, Benjamin Rouben, Matthew Fountain,

Steve Feldman, Ronald Raines, Elliott Roberts,

. and the proposer, J. Weatherlv.

N/D 2. Find a number thai equals its own logarithm.

There is no positive real number x such thateither logl0(x) = x or log^x) = x. Several readers,

including the proposer. Smith D. Turner (/dt),

went on to consider log,. However. I do not feel

this meets the conditions of the problem (note

that logarithm was used in the singular). Tim Ma-

loney (and others) made another generalization;he writes:

First, one must recognize that the solution is a

complete number, call it Z = u + iw = re"Then

re* = ln(re*)

r-eosO + irsin 0 = In (r) + i 6, or

ln(rVr = cos 8

r = 0/sinO.

We must therefore solve

In(6/sin8) » 8-cot fl.

An iterative solution gives

8 = 1.337236 .... so

Z = (1.374557 . . .)e": 5)r2* '

= (.3181313 . . .) + 1(1.337236 . .).

I must admit that I first saw this problem in my

junir.: year at M.l.T. (1970), when someone pro-

post ' the problem in a lunchtime discussion in

Protessor Daniel Kleppner's research group.

Klenpner immediately drew a graph on the board

to ove the solution could not be real, asserted

that it must be complex, and left us all speech

less.

Also solved by Eric Rayboy, R. Bart, Matthew

Fountain, Ronald Raines, Winslow Hartford, John

Spalding, John Woolston, Naomi Markovitz, Mike

DebesCaip.

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Hennessey, Frank Carbin, Greg Huber, Avi Orn-

stein, Oren Cheyette. John Trussing. Edwin

McMillan, Charles Sullon, and the proposer.

Smith D. Turner (/dt).

N/D 3. Given a door lamp wilh two bulbs, in

which each socket has a chain which when pulled

will change the on/off stale of only the bulb in

that socket. When the lamp is on. it is difficult lo

determine whether one (and if so which one) or

both bulbs are on. The problem is to find the

shortest sequence of pulls that will turn the lamp

completely off sometime during the sequence

(e.g. if chains arc labeled A and B, AAB0AB fits

the requirements, but a shorter sequence may be

found). Can you generalize your solution to a

lamp of three bulbs, four bulbs, etc.? Is your solu

tion unique?

The following solution is from John Spalding:

The solutions can be generated recursively

starting with the solution for the trivial case of

one bulb. Suppose we number the bulbs 1, 2, 3

. . . instead of lettering them. Then the solution

for n bulbs could be given as the following func

tion f(n) giving a sequence of numbers:

f(D = 1f(n) = f(n - 1), n, f(n - 1)

For two bulbs, the solution is thus 1.2,1 or ABA:

for three bulbs, the solution generated would be

ABACABA; for four, ABACABADABACABA, etc.

When the chains are pulled in this sequence, the

bulbs will cycle through what I think is called a

Gray Code, in which all the possible n-digit bi

nary numbers are generated by only changing

one digit at a time. The number of steps required

for n bulbs is thus 2" - 1, which is what we

would expect. Is the solution unique? Well, no.

Multiple working sequences may be generated by

simply permuting the bulbs' labels—e.g., ABA

and BAB. In addition, labels may be permuted al

intermediate levels of the recursion—e.g., ABA

CABA and ABACBAB are working sequences for

three bulbs.


Fountain, Winslow Hartiord, John Woolslon, Na

omi Markovitz, Mike Hennessey, Frank Carbin,

Oren Cheyette, Pat Kinney, Mike Strieby, Harry

Zaremba, Howard Stem, Yokichi Tamaka, and

Joe Feil.

N/D 4. A regular hexagon can be inscribed in an

equilateral triangle so that its alternate sides coin

cide with the sides of the triangle. What is the ra

tio between the areas of the hexagon and the

triangle?

Walter Cluett has little trouble wilh this one:

Divide the hexagon into six equilateral triangles

and the answer is 6 to 9 or 1 lo 1 1/2.

Also solved by Eric Rayboy, R. Bart, Steve Feld-man, Howard Stern, Ronald Raines. Winslow

Hartford, John Spalding, John Woolston, Naomi

Markovitz. Mike Hennessey, Frank Carbin. Greg

Huber, Avi Ornslein, Pal Kinney, Harry Zar

emba, Frederic Jelen, Ruben Cohen. Stefan Habs-

burg, Peter Silverberg, George Byrd, James

Reswick, Dick Robnett, Raymond Gaillard. Mi

chael Lamoureux. Man' Lindenberg, Smith D

Turner (/dt), and the proposer, Phelps Meaker.

N/D 5. In the country of Moolah, the national

bank issues new dola bills to replace each bill that

wears out or is lost or destroyed, so there is al

ways a constant number of dolas in circulation.

On January 1, the bank issued a new bill with the

picture of Prince Centime replacing that of the

late Queen Peseta. After one year, they found

that 10/27 of the bills in circulation were the new

variety. After two years, 2/3 of the bills; and after

four years all the bills were the new type. What is

the life expectancy of a dola bill?

Harry Zaremba sent us a lucid solution:

Assume N to be constant number of bills in circu

lation. In terms of N and their common denomi

nator, the fractional amounts of new bills in

circulation during the successive four years were

ION/27, 18N/27, 24N/27, and 27N/27. The number

of old-variety bills that were replaced in each of

the four years was ION/27, 8NV27, 6N/27, and 3N/

27, and their respective average years in circulation

were 1/2, 3/2, 5/2, and 7/2 years per dola. Let A

be the weighted average life expectancy' of the old

bills. Then,

AN = ION/271/2 + 8N/27-3/2 + 6N/27-5/2 +

3N/27-7/2, or

A = 1.574 years per dolla.


Fountain, Winslow Hartford, Frederic Jelen, and

the proposer, Frank Rubin.


Y 1984. Claes Wihlborg and Mats Ohlin have re

sponded.

M/J 1. R. Bart found two alternative solutions.

JUL 3. Smith D. Turner (/dt) found a simpler so

lution strategy.

A/S 1. Samuel Levitin and Benjamin Rouben have

responded.

A/S 2. Samuel Levitin has responded.

OCT 1. R. Bart, W. Smith, and Richard Hess

have responded.

OCT 2. R. Bart has responded.

OCT 3. R. Bart, Samuel Levitin, Richard Hess,

and R. Morgan have responded.

OCT 4. R. Bart, Richard Hess, Samuel Levitin,

Phelps Meaker, and Altamash Kamal have re

sponded.

OCT 5. R. Bart has responded.

N/D SD 1. Michael Strieby and Dick Robnetl

found alternate solutions.

N/D SD 2. The correct answer is 8, as noted by

Eric Rayboy, David DeLeeuw, R. Bart, Winslow

Hartford, John Spalding, John Woolston, Naomi

Markovitz, Pat Kinney, Mike Hennessey, Mike

Strieby, Frederic Jelen, Ruben Cohen, Stefan

Habsburg, Peter Silverberg, George Byrd, James

Reswick, Dick Robnett, Mike Ober, Walter Cluett.

Roger Allen, Harry "Hap" Hazard, Peter DeFoe.

Victor Christiansen, W. Katz, and Phelps Meaker.


SD 1. East has 12 cards in the majors, most likely

6-6-1-0. She needs the VA for her opening bid. In

order to get back to your hand to draw the last

trump, you must not play a spade! (1) It may be

trumped (true!), and (2) You need the *A to exe

cute a so-called "simple" squeeze play. (What

squeeze is simple al the table?). Ruff a heart and

play all your trumps but one, reaching this four-

card position:

North:

* K85

South:

* A93

A3

East:

*QJ10

V A

After you play your last club, discarding a spade

from dummy. East is rendered helpless.

SD 2. The egg. The first egg from which a

chicken hatched had to have been laid by the im

mediate evolutionary ancestor of the chicken. On

the other hand, the first chicken laid something

like what we have for breakfast—what else, a dinosaur egg? Please forward all correspondence to

the Nobel Institute in Stockholm, Sweden. 1'U bewaiting for it there.

A32 APRIL 1985

PUZZLE CORNER

ALLAN J. GOTTLIEB

How to Computerize Your Engagements

As promised in the October issue, here

is the current backlog of submitted prob

lems in various classes. The shortest de

lay until publication is for speed

problems, where I have a half-year sup

ply. The backlogs for chess, bridge, and

regular problems are each approxi

mately one year. Some confusion has

developed concerning the computer-re

lated problems. These problems will be

clearly identified when presented, and

I did not intend to suggest a preference

for computer calculations over mathe

matical analyses for regular problems.

To date (November 9), I have received

two problems designed as computer-re

lated, one of which appears as F/M 1

below. Thus, the backlog for this class

is just one problem.

Finally, I must apologize to Merle

Smith for misspelling his name in the

October column.

Problems

F/M 1. Alfred Anderson inaugurates the

computer-oriented problems with this

offering; he writes:

Recently I used brute force to solve a

rather interesting computer-oriented

problem. Perhaps one of your readers

would find a more elegant solution.

Management meetings are scheduled on

the second Thursday of each month, ad

ministrative conferences are the third

Friday, and work units have a seminar

on the first Monday. Derive an algo

rithm which will generate a date given

the year, month, day-of-week, and or

dinal week within the month. For ex

ample, if a meeting were scheduled for

the third Friday of August 1984, the al

gorithm would return "August 17."

Note that a meeting on the fifth Tuesday

in March would be fine for 1983, 1985,

and 1986 but not for 1984 (there are only



GOTTLIEB. '67. ASSOCIATE




NEW YORK UNIVERSITY, 251


N.Y.. 10012.

four Tuesdays in March of 1984). In this

case the algorithm could return January

0.

F/M 2. William Stein likes to deal with

loosely coupled coins:

Two coins, loosely coupled, are

flipped simultaneously such that ifeither one is heads, the other has prob

ability 7/8 of also being heads, but if any

one is tails, the other is equally likely to

be either heads or tails. Find the prob

ability of each individual coin turning up

heads, and the probability of their both

being heads simultaneously (or prove

that the problem statement and data are

inconsistent).

F/M 3. A geometry problem from Phelps

Meaker:

A horizontal line of length 2a forms

the common base for two isosceles tri

angles. On the near side the triangle is

45° - 45° - 90°, and on the opposite side

75° -75° -30°. Determine the radius of

the circle tangent to all sides of the com

posite lanceolate figure, and locate its

center.

F/M 4. Smith D. Turner J(dt) wants you

to find a four-digit number whose

square is an eight-digit number whose

middle four digits are zero.

F/M 5. David Dreyfuss was attracted to

the following problem:

Consider two dipoles with dimen

sions as indicated. The lower dipole is

fixed, and the upper dipole is con

strained to move along a horizontal line.

(This is roughly the geometry encoun

tered in magnetic stirring.) Find the con

ditions on h and b for which the upper

dipole tends to center (the force is in the

opposite direction of d, the displace

ment from the center line). When does

the motion of the upper dipole approx

imate simple harmonic motion?

Speed Department

SD 1. David Evans has placed a turtle

in each of the four corners of a square

room measuring 3 meters on a side. All

four start moving.at the same instant at

a constant speed of 1 cm./sec, and each

crawls directly toward the turtle to the

left. How long does it take for them to

meet at the center of the room?

SD 2. Steven Bernstein knows a teacher

who brings apples for her students:

Ms. Lang, the third-grade teacher,

wanted to do something nice for the n

students in her class. One day she

brought in an apple which would be

given to a lucky youngster. Her problem

was how to choose the lucky one fairly.

Here is what she decided to do: She said:

"I'll pick a number from 1 to n. The first

one of you to guess the number is the

lucky winner. Let's hear your guesses

in alphabetical order. Aaron, you're

first. Be sure to speak up so everyone

can hear you. Zelda, you'll be last."

"Unfair!", said Aaron. "I'll have all

those numbers to choose from. My

chances of guessing are pretty small.

Wanda, Yolanda, and Zelda will have a

chance to hear everyone else's guesses

so they will have a better chance of win

ning!" "Wait a minute," said Zelda. "1

probably won't even get a chance to

guess because all the other kids go first

and one of them will win before it's even

my turn!" Ms. Lang replied, "Children,

I've thought about it and this procedure

is fair. You all have the same chance of

winning." Is Ms. Lang correct?

Solutions

OCT 1. What is the minimum number of high-card points needed to make a contract of 7 spades

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HARVARD

COOPERATIVESOCIETY

7Z£"*' Vlsa

against the best defense?

Bob Sackheim sent us the following:

I suspect that the minimum number of high-

card points needed to make a 7-spade contract is

five. One possible deal would be as follows (point

cards are shown; all other cards are indicated by

x's, their values being immaterial):

6 x x x x x

A K

¥ AQxx

♦ K

* AKxxxxx

♦ xxxxxxxx

*-

A A ) x x x x

Vxxxxxxx

♦ Q¥ KJ

♦ AQJx

+ QJxx xx"

If West leads the AK, South takes with the A A,

cross-ruffs four hearts and four diamonds, setting

up three hearts in his hand, plus his last spade

for 13 tricks. If West leads a heart. North ruffs,

leads a spade to the A A, then three more hearts

and three more diamonds are cross-ruffed, setting

up three good hearts in South's hand, plus his

two remaining spades for 13 tricks. If West leads

a diamond. South ruffs, plays the AA, then

cross-ruffs four hearts and three diamonds, end

ing in dummy; his last spade, and either three

hearts or three diamonds, are good. If West leads

a club. South ruffs, plays his AA, cross-ruffs four

hearts and four diamonds, ending in his hand;

his last three hearts are good.

Also solved by Tom Harriman, Lester Steffens,

Richard Boulay, Philip Dangel, Alan Robock,

Warren Himmelberger, Winslow Hartford, and

the proposer, Howard Sard.

OCT 2. Given a rotating hollow semi-sphere with

a hole in the bottom, into which a marble is

dropped, find the angular velocity required to

hold the marble exactly halfway between top and

bottom of the sphere. Assume the frictional coef

ficient between marble and sphere is |i and the

radius of the sphere is R.

R m

rsinS

In figure shown,

r = radius of the marble

R = radius of the hemisphere

W = weight of the marble

w = angular velocity of the hemisphere

g - acceleration of gravity

and A is the point of contact between the marble

and the spherical surface at a height R/2 above

the bottom. For equilibrium in a vertical uni

formly rotating plane, the moment of W about

point A will equal the moment of the centrifugal

inertia force W/gy/hc about the same point. Thus,Wr sin 8 = W/g-wVr cos 6, or

i 6. Since

9 = cos-1(r/2)/R = 60°,

then tan B = VJ. Alsax = (R - r)sin 6 = V3/2-(R - r).6ubstitutinfi x and tan 8 into equation (1) yields

w = V2g/(R - r) rads./sec., the required angular

velocity. If the intent of the problem was to have

the center of the marble halfway above the bot

tom, we would have

cos 8 = R/[2(R - r)] and

sin 8 = x/(R — r). Hence, for this condition, the

angular velocity from (1) would be:

w = V2g/R rads./sec.

Also solved by Matthew Fountain.

OCT 3. In the Illinois Lottery Lotto game,-the

player chooses six different integers from 1 to 40.If the six match, in any order, the six different in

tegers drawn by the lottery, the player wins the

grand prize jackpot which starts at SI million and

grows weekly until won. Multiple winners split

the pot equally. For each SI bet, the player must

pick two, presumably different, sets of six inte

gers. Considering the grand prize alone, under

what conditions would it pay, on the average, to

play this game? In the game week ending June

18, 1983, 78 people matched all six winning inte

gers and split the jackpot. Estimate the odds of

this outcome, given that 2 million people bought

SI tickets that week.

There are f ^ I or 3,838,380 possible combinations

which might be chosen. It will prove highly con

venient to round this off to 4 x 10s. The expecta

tion for a lottery participant would be easy to

compute, if we didn't have to worry about multi

ple winners. Choosing 2 out of 4,000,000 combi

nations, someone would have a l/2,000,000th

chance of winning. The prize would have to ex

ceed S2 million before it would pay, on the aver

age, to play the game. But, in fact, the prize

would have to be significantly higher than this,

because there is a good chance of multiple win

ners. The probability that a winner will have to

share his prize is only 1/e, as can easily be

shown. There are, according to information given

later in the problem, 2,000,000 people playing the

game. That is, 4,000,000 combinations are se

lected. Since there are 4,000,000 possible combina

tions, the probability that any given selection

does not match the winner (assuming that each se

lection is equally likely) is '

|(4 x 106) - l]/(4 x 10s).

The probability that all the other selections don't

match the winner is(|(4 x 10s) - l)/4 x 10*)« x 10s.

This is very dose to 1/e since, as is familiar from

calculus,

lim ((n - l)/nj" = 1/e.

So there is a probability of 1 - 1/e that there will

be additional winners. We will need to know the

probabilities of specific numbers of additional

winners, and these can be obtained from the

I'oisson formula:

y

1/e).37 (

.37

.18

.06

.01

negligible

i h

This gives the probability of k additional winners,

where u. is the average number of winners. In the

present case |i is 1, since there are 2,000,000 par

ticipants, each with a l/2,000,000th chance of suc

cess. The following table may be compiled:

Probability

No additional winner 37 (1)

One additional winner

Two additional winners

Three additional winners

Four additional winners

Five or more winners gg

What this indicates is that a winner has a .37

chance of getting all the money, a .37 chance of

getting half the money, etc. What the winner

could expect would be, roughly,

.34 + .37/2 + .18/3 + .06/4 + .01/5 = .63

of the prize money. In order for the lottery to be

a good bet, 63/100 times 1/2,000,000 times the

prize money would have to exceed SI. That is,

the prize money would have to exceed S3.17 mil

lion.

To determine the probability of the June 18,

1983, result, we again make use of the Poisson

formula, with k = 78. This gives the probability

that 78 people will have a given winning combi

nation. Since 78! is on the order of 10'", this is anextremely low probability. An event this unlikely

could not, practically speaking, have happened.

But it did happen. Therefore, something has gone

wrong in the preceding calculations.

The problem lies in the assumption that each

combination is equally likely to be picked by lot

tery participants. In fact, people do not choose

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numbers at random. They choose numbers which

have some significance. Often this is a strictly in

dividual significance (one's age, a date of the

month with special meaning, etc.). But the

choices are sometimes on the basis of a more gen

eral significance. Some numbers (e.g., 7 and, to a

lesser extent, 3) are believed to be lucky. These

and related numbers, such as 21, 33, and 37, will

be heavily over-represented in the selections. By a

reverse psychology, 13 might also be popular.

What were the winning numbers for June 18,

1983? This can be looked up in daily papers of the

region for June 19. When I did look this up, I had

some idea, based on the considerations in the last

paragraph, of what 1 might be seeing. But, even

so, I could hardly believe my eyes. The winning

combination was: 7, 13, 14, 21, 28 and 33. It

would be difficult to imagine a group more ap

pealing to the believer in lucky numbers! Here we

have all the products of 7 between 1 and 40, with

13 thrown in for good measure.

So the answer to the question "What was the

likelihood of the June 18, 1983, result?" is now

apparent, though it doesn't lend itself to exact

quantification. It was quite remarkable (having a

probability of 1/(4 x 10s) that that particular com

bination was the winner. It was not at all remarka

ble that 78 people chose it. The probability of the

complex event (that number winning and being

chosen by 78) is not much less than 1/(4 x 10°).

Let's consider the more general question of

there being some result involving a very large

number (say 50 or more) of winners. I am in

clined to think that this is not extraordinary un

likely. Take the following as a list (obviously not

exhaustive or definitive) of lucky numbers be

tween 1 and 40: 3, 7, 11, 13. 14, 21, 22. 28, 33, 35.

No other winners

One other winner

Two other winners

Three other winners

Probability

.61

.30

.08

.01

37. There are 462 combinations of these

numbers which might be selected. Given that the

total of possibilities is 4 x 10s, it's not likely thai

any of these will again be selected in the foresee

able lifespan of the lottery. But consider the com

binations of five of these numbers with some

other number selected for its significance to an in

dividual. There are (™\ x 29 (about 14,400) ofthese combinations. The chance that one of them

will be a winner is roughly .004 (1 in 250). So one

of these should come up about every five years,

and when it does there may well be a lot of win

ners. Probably not as many as 78, for these com

binations don't have the unique appeal of that

other one, but 50 would not seem too unreasona

ble for some of these combinations.

Ifs now quite clear that my earlier calculation

of what percentage of the prize a lottery player

might expect to collect can only be interpreted as

an average. What a particular player can expect

will vary in relation to the numbers that are se

lected. If he picks two of the supposedly lucky

combinations, his winnings may have to be

shared with 50 others. If the grand prize is at the

$1 million level, each winner's share is slightly

less than $20,000. Recalling that there is still just a

1/2,000,000th chance of winning, this works out

to an expectation of less than \t.

If you deliberately choose nothing but "ordi

nary" numbers, your expectation is somewhat im

proved beyond the level I gave earlier. Let's try a

quite arbitrary assumption, in order to make the

discussion more concrete. Suppose half of the lot

tery entrants make use (consciously) of one or

more of the "lucky" numbers in picking their

combinations. The other half choose their num

bers on the basis of considerations peculiar to

themselves, so that these selections are, in the ag

gregate, random. If you deliberately choose only

"ordinary" numbers, then you are competiting

only with the second group. There is no possibil

ity that you will have to share your prize with

anyone in the first group. Using the Poisson for

mula, we can compute a table of probabilities for

this situation, similar to the table given earlier.

But now |i is just 0.5. We have, on the supposi

tion that you win:

You would stand to receive .61 + .30/2 + .08/3

+ .01/4, or about .79 of the prize. In the case

where the prize is SI million, this works out to an

expectation of 40j (compared to the 1« expecta

tion enjoyed by the other contestant). But note

that under any circumstances, the prize must be

significantly above $2 million before the lottery

becomes a good bet.

Also solved by Frank Carbin, Warren Himmel-

berger, Matthew Fountain, and the proposer. Jon

athan Hardis, who sent us a particularly complete

solution. Mr. Hardis seems to be quite an author

ity on the Illinois lottery. Since he has a Chicago

address, one might conjecture that some of his

knowledge comes from empirical study.

OCT 4. An associate research professor walks

into his office one morning and says to his secre

tary, "I had three dinner guests last night. The

product of their ages was 2450. The sum of their

ages was twice your age. Can you tell me their

three ages?" Ten minutes later his secretary came

to him and said that the problem could not be

solved. He said, "You are right. I will now tell

you that I was the oldest one there." The secre

tary was then able to tell him the ages of.the

three dinner guests. What are the ages of the din

ner guests, her age, and the professor's age?

The following solution is from Tom Gallahan:

The first thing to do is make a list of the possible

combinations of ages. Once the secretary has

done this he/she can eliminate the ones that do

not add up to twice his/her age. If there were

only one set of ages that fit the criteria it would

be the answer. This is not the case because he/she

cannot solve the problem. There must be more

than one choice. The fact that the professor is the

oldest one there must distinguish between the

possible choices so that the secretary can solve

the problem.

In making a list it helps to make a factor tree,

since three ages's product is 2450.

2450

\98

2 49

/\

25

/ \5 5

2450 = 2x5x5x7x7 7 7

Ages of Resulting age

guests of secretary

1 2, 5,245 126

2 2, 7, 175 92

3 2, 25,49 38

4 2, 35, 35 36

5 5,5,98 54

6 5, 7, 70 41

7 5, 10,49 32

8 5. 14,35 27

9 7, 7, 50 32

10 7, 10,35 26

11 7, 14, 25 23

You must also realize that one or two of the

guests may be one year old:

12 1,25,98 62

13 1,35,70 53

14 1,49,50 50

15 1, 1, 2450 1226

None of the guests may be 0 years old because

the product of the guests' ages in that case will

always be 0.

For ease of explanation 1 have numbered the

sets of ages. All of the resulting secretary's ages

are distinct except 7 and 9. One of these must be

the correct answer; thus we now know that the

secretary is 32 years old. In case 7 the professor

must be 50 years old or older to be the oldest one

there. In case 9 he must be 51 or older. If the pro

fessor is 51 or older the secretary can't choose be

tween the sets. The professor must be 50 years

Ml FEBRUARY/MARCH 1985

old and 7 must be Ihe correct set. logically we

could have eliminated 1, 2, and 15 but this is un

necessary.

Also solved by Michael Jung, Matthew Foun

tain, David Griesedieck, Tom Harriman, Naomi

Markovitz, Richard Boulay, Fernando Saldanha,

Jerry Sheldon, Clarence Cantor, Steve Feldman,

Dennis Loring, Ronald Raines, James Michelman,

Avi Omstcin, Tso Yce Fan, John Rosendahl, Ray

mond Caillard, E.R. Foster, Danny Mintz, Roy

Levitch, Miriam Nadel, Tom Lydon, Myles Fried

man. Pierre Heftier, Frank Carbin, VVinslow Hart

ford, Harry Zaremba, and the proposer. Merle

Smith.

OCT 5. Given an irregular polygon of n sides, in

which sequence should Ihe sides be arranged and

how should the comer angles be determined to

give the greatest area?

The following solution is from the proposer,

Irving Hopkins:

Consider that the polygon is the horizontal

cross-section of a vertical open-topped water

slandpide, the rectangular sides of which are

hinged together along their vertical edges. There

is no friction in the hinges, and Ihe lower end of

the vessel rests in a friction-free manner on a hor

izontal plate with no leakage of water. Gravity

causes the water to fall until the sides have

moved to what must be the enclosure of maxi

mum area. When movement of the water has

ceased, the pressure at any depth is uniform in

all directions. The stress in the sides depends on

the depth, but the geometry does not vary with

the pressure. Assume a depth at which a side of

width L and a suitable vertical dimension is sub

ject to a force 2L. At the junction of L, and L2 the

forces are as shown, where t, is the tension in L,

and t: that in L:.

Length L, l|

At the hingeptn, Ihe forces parallel to L, balance

if

t, = tjCOsA, + L2sinA, (1)

and the perpendicular forces if

L, + LjCosA, = t;SinA, (2)

Going on to other comers, we have:

tj = t)COsA2 + L>sinA2 (3)

Lj + LjCOsA2 = t}sinAj (4)

(2n - 1)t. = tgCOsA, + L,sinA,,

L, + L]CosA, = t|SinA,. (n)

From the even-numbered equations above:

t, = (L, + L,cosAJ/sinA.

t2 = (L, -t- LjCosA,).'sinA,

t, = (L,_, + UcosA, - l)/sinA,.

Substituting these values of t in the odd-num

bered equations, we get:

(L,cosA. + L,)/sinA. = (L,cosA, + LJ/sinA, (I)

(L^osA) + Li)/sinAi = (L2cosA2 + Ls)/sinA2

(H)

(UcosA,., + L,.|)/sin A,,., = (L,cosA, + Lt)/

sinA,. (Ill)

We now have n equations from which to find the

values of angles A, to A.. But there is one more

requirement: the sum of Ai to A, must be 360s.

Assume a value for A] and let Q equal the left-

hand side of (IT). Square both sides of (IT), which

becomes

cosJAi(L,! + Q2) + 2LjLjCosA2 + (Lj2 - Q2) =O.

This is a quadratic equation in cosA? from which

A] may be found, and so on. For a pentagon with

sides 3, 5, 7, 9 and 11 in length the angles were

found to be

Goldbeig-Zoino &Associates Inc

Gaolochnlcnl-

Goohydrologlcal

Consultants

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EngineeringSite Planning ft

Dovolopmont

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f. W. Clark, '79

W. E. Hodge, 79

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C. A. Llndberg. 78

R. M. Simon, 72

E. I. Stolnberg, '60

T. vonRosenvlnge IV, '80

The GEO Building

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government and

developers of high


lor over 35 years

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ARCHITECTSENGINEERSpi &MKIPQQ


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to tho nation slnco

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cttlos throughout tho

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Alexander W. MoHat. Jr.

H.H.

Hawkins&SansCompany

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Hlnglum, MA 02043

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(617) 749-6012

Koalth Care Consultant*


Management

Subsidiaries:

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Assoc. Inc.


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Corporation

Rockford Convalescent

Center Inc.

Chambra Corporation

Chariot N. Oobes '35

5666 Strathmore Drive

Rockford, IL. 61107

75916367

37 590272"

57.838328

68.359986° X ,2396886.

80 4 25299"

108.22973"

A,

A2

A,

A,

A,

37.590°

57.838°

80.425°

108.230"

75.916°

Could such polygons be circumscribed by circles?

If so, each side of the polygon is a chord of the

circle, whose center must lie at the point of inter

section of the perpendicular bisectors of the

chords. Consider two adjacent sides of the poly

gons, say L, and Lj, with angle A,.

Taking

c = (LV2)sinA, + qcosA, and horizontally

L,/2 = qsinA, - (Lj/2)cosA,.

From these equations we find that

c = [Lj/2 + (L|/2)cosA,]/sinA,,

and the radius R12 determined from sides L, andLjis

R!2 = (c3 + (L,^)2)"2

|(j) (.ajcos

which boils down to

4R»2 = (L,2 + 2L,LjCOsA, + L22)/sin2A, (IV)The existence of a circumscribing circle depends

on R,2 = Ra = . . R.,. By analogy with (IV),

4Ra! = (Lj2 + 2L2LjCosA2 + L32)/sin2A2 (V)Squaring both sides of (11), we get

(L,2 + 2L,L2cosA, + LjWA,)/sin2A, = (L,2 +2LjL3cosA2 + L^cos^j/sin'Aj. (VI)Subtracting the left side of (VI) from the right sideof (IV), and the right side of (VI) from the right

side of (V), we get

LjJ(l - cos2A,)/sin2A, = L,2, and

L22(l - cos2A2)'sin2A2 = W-The right sides of (IV) and (V) are therefore equal;

hence RI2 and Ro and all the other R's are equal.

The area of each triangle is easily found by

Area = (L/2)[R2 - (L/2)2]"3. The sequence inwhich the sides are arranged is immaterial, equiv

alent to carelessly cutting a pie and then shuffling

the pieces.

The pentagon described is shown above.

Each side subtends an angle at the center [B, =

2arcsin(L/2R)), and the sum of these must equal

360°. The easiest way to find the radius giving B

= 360° is cut-and-try. This may fail if a very long

side, U, causes the center of the circumscribing

circle to be outside the enclosure. In this case,

solve for

Also solved by Matthew Fountain, Tom Ham-

man, Winslow Hartford, and Harry Zaremba.


M/J 4. Andre Schmitz found a simpler way to

present the solution.

A/S 4. Dick Swenson has responded.


SD 1. Five minutes. Since the turtles move at

right angles to each other, an approached turtle's

motion does not contribute to the distance the approaching turtle must travel.

SD 2. Yes, Ms. Lang is correct. Each child has

probability 1/n of winning. Aaron's chance is 1,'n

because he has n numbers to choose from. The k-

th child will win if each of the k - 1 children that

went before him/her guessed incorrectly and the

k-th guesses correctly. The probability of this happening is:

[(n - l)/n|[(n - 2)/(n - 1)] . . . (n - k + l)/(n- k + 2)[l/(n - k + 1)] = I/n.

Notice that in order for this to work it is neces

sary for each guess to be heard by the children to

follow. In this way the disadvantage of going

near the end is exactly compensated by a narrow

ing down of choices.

A24 FEBRUARY/MARCH 1985

Ct ' 1 1

Goip.


development In

radio frequency,

mlcrowovo and

mllimeter wavo

engineering and

rotatod areas.

RF and Microwave

Systems Design


of Microwave Power

Precision

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ier

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system uses GPS

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per-mllllon accuracy In

all three coordinates.

PUZZLE CORNER

ALLAN J. GOTTLIEB

100 Ways to Say 1985

BoyleErigineerijGap.

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238 Main St.Cambridge. MA 02142

(617) 492-6460

Thomas S. Maddock '51

1501 Quail Street

P.O. Boi 7350

Newport Beach, CA

92660

(714) 752-1330

Harl P. Aldrlch, Jr. '47


Edward B. Klnrtar '67

Douglas G. Gifford '71

Joseph J. Rlxner '68

John P. Dugan '68

Kenneth L. Reckor '73

Mark X. Haley °75

Robin B. Dill '77

Andrew F. McKown '78

Keith E. Johnson 'B0

This being the first issue of another

year, we again offer a "yearly

problem" in which you are to ex

press small integers in terms of the digits

of the new year (1, 9, 8, and 5) and the

arithmetic operators. The problem is for

mally stated in the "Problems" section,

and the solution to the 1984 yearly problem is in the "Solutions" section.

Problems

Y1985. Form as many as possible of the

integers from 1 to 100 using the digits

1, 9, 8, and 5 exactly once each and the

operators +, -, x (multiplication),/(di-

vision), and exponentiation. We desire

solutions containing the minimum num

ber of operators; and, among solutions

having the minimum number of oper

ators, those using the digits in the order

1, 9, 8, and 5 are preferred. Parentheses

may be used for grouping; they do not

count as operators.

JAN 1. Our next problem is the last

member of Emmet Duffy's collection of

seven-card bridge problems. For the cur

rent challenge, South is on lead and is

to take six tricks against best defense

with hearts as trump:

A

V

♦

J8

J10 9 8

5

*

♦

♦

A

V

♦

*

983

10

AQ

AK—

—

AQ

2

7

43

A

V

♦

10

6

K6

KJ6

JAN 2. Bruce Calder, after working on

1983 N/D 4, sent us the following spin-

off, a problem demonstrating the ele

gant subtlety of Newtonian mechanics:

A smooth, rigid, and circular wire

SEND PROBLEMS, SOLUTIONS.

AND COMMENTS TO ALLAN /

GOTTLIEB. 67. ASSOCIATE




NEW YORK UNIVERSITY, 251


NY.. 10012.

hoop hangs from a rigid support by an

ideal, extensionless string. Two small

beads slide along the hoop (like beads

of a necklace) with negligible drag and

friction. The beads are slid to the top of

the hoop and released. How massive

must each bead be to spontaneously lift

the hoop?

JAN 3. Here's one John Rule dug out of

the file where he keeps interesting prob

lems encountered from various sources:

A man received a check calling for a

certain amount of money in dollars and

cents. When he went to cash the check,

the teller made a mistake and paid him

the amount which was written in cents

in dollars, and vice-versa. Later, after

$3.50, the man suddenly re

alized that he had twice the amount of

money the checked called for. What was

the amount on the check?

JAN 4. Our last regular problem is from

Floyd Kosch:

yyy

MA JANUARY I

Frank E. Reeves, '24; August 26. 1984; 1661 Mo

hawk St., Los Angeles, Calif.

Lloyd R. Rogers, '24; December 31, 1984; 2129 Pol

Springs Rd., Lulherville, Md.

Gavin Watson, '24; September 21, 1984; PO Box

1099, Arizona Bank, c/oTrust Dept., Sun City, Ariz.

Francis E. Field, '25; July 6, 1984; 32 Buena Vista

Rd., Biltmore Forest, Asheville, N.C.

Harry Newman, '25; July 1984. 220 W Jersey St.,

Elizabeth, NJ.

Robert A. Nisbct, '26; July 21, 1984; 25 Barrows

Terr., Stratford, Conn.

Robert N.C. Hesscl, '27; July 20, 1934; 22 Saxib Rd.,

Worcester, Mass.

Cordon E. Thomas, '27; September 1984; 20 Terrene

Ave., Nan'ck, Mass.

Victor J. Decode, '28; July 9, 1984; Ocean Club No.

1109, 4020 Gait Ocean Dr., Fort Lauderdale, Fla.

Prescott D. Crout, '29; September 25, 1984; 9 Pine-

wood St., Lexington, Mass.

Richard V. Does, '29; 1984; 27 Hamilton Ave., Ded-

ham. Mass.

Emmette F. fcard, '29; September 3, 1984; RT 1 Box

159, Hazlehurst, Miss.

Delmer S. Fahrney, '30; September 12, 1984; 10245

Vivera Dr., La Mesa, Calif.

Charles G. Habley, '30; September 8,1984; 210 Mid-

vale St., c/o Felker, Falls Church, Va.

Ellas Klein, '30; January 17,1984; PO Box 115, Half

way House, Transvaal, S Africa.

Harry W. Poole, '30; September 7, 1984; 1201 Mot-

tron Dr., McLean, Va.

Reuben Roseman, '30; March 1984; 520 Wyngate

Rd., Timonium, Md.

Glenn Goodhand, '31; September 11. 1984; 6307

Stoneham Ln., McLean, Va.

Ernest H. Lyons, Jr., '31; 1984; 23871 Willows Dr.

No. 350, Laguna Hills, Calif.

Heinrich W. Weitz, '31; August 28, 1981.

Lucien B. Curtis, '32; 1984; 24 Park St., Brandon,

Vt.

Edwin Allen Newcomb, '32; September 14, 1984;

Stillwater Health Care, Stillwater Ave., Bangor,

Maine.

Gerner A. Olsen, '32; April 26, 1984; 5 B 7 Glen

Ave., Scotia, N.Y.

Frank Der Yuen, '33; July 23, 1984; 1565 Kalaniiki

St., Honolulu, Hawaii.

Gordon A. Danforth, '34; September 19, 1984; PO

Box 444, Richmond, 111.

George E. Agnew, '35; January 3, 1984; 17881 Corte

Emparrado, San Diego, Calif.

Robert Kenneth Bullington, '37; May 30. 1984; 36

Glenwood Rd., Colts Neck, N.J.

Robert D. Morton, '37; October 14, 1984; 82 Sunset

Farm Rd., West Hartford, Conn.

Carl I. Shulman, '38; October 15, 1984; 41 Bethune,

New York, N.Y.

Philip W. Constance, '39; March 31, 1984.

David B. Parker, '40; March 30. 1980.

Bjom Lund, '41; September 12, 1984; 194 Niantic

River Rd., Waterford, Conn.

Neil D. Cogan, '42; August 29, 1984; 47 So Ridge-

land Rd., Wallingford. Conn.

Andrew L. Johnson, '43; October 13, 1984; 211 En-

glewood Ave., Newcastle, Penn.

Donald T. Cloke, '45; July 21,1984; PO Box 173. La

Grange, Maine.

Arthur C. Nisula, '46; September 3, 1984; 907 Ma

rina Dr. No. 401, North Palm Beach, Fla.

Elmer B. Sampson, '47; September 7, 1984; RR 3Box 1368, Leesburg, Fla.

Stanley H. Southwick, '51; April 20. 1984; 935 Ma

ple St., Friend, Neb.

James W. Burch, '53; April 10, 1984; 6 Taney Ave.,

Annapolis, Md.

John M. Houston, '55; August 31, 1984; 1302 RoweRd., Schenectady, N.Y.

Robert D. Alter, '56; March 21, 1984; 258 Woodland

Rd., Highland Park, 111.

Harvey J. Baker, '72; September 9,1984; 3122 North

ampton St. NW, Washington. D.C.

Otto K. Soulavy Burchard, '83; September 1984;

Calle Caurimare, Res Parque, Colinas De Bello

Monte No. 62A, Caracas, Venezuela.

R. Nicolas Harrison, '84; October 6, 1984; 24 Elm

Grove Rd., Ealing London. England.

Edward R.Mantel

Gup.

Builders lor Industry,



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technology facllltias

lor over 35 years


Konnoth fl. Hoffman 78

Oouglas R. Marden '82

280 Lincoln Stroet

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Raman &Associates Inc.

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Institutional One Qateway Center

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Now York, N.Y.

10038

840 Memorial Dr.

Cambridge, MA

02139

575 Mission St.

San Francisco, CA

94105

Goldbeig-

Zaino &Associates Inc.

Geotechnlcal- D. T. Goldberg, '54

Goohydraloglcal W. S. Zolno, 'S4

Consultant! J. D. Guortln, '67

Foundation/Sell M. J. Barvenik, '76*

Engineering M. D. Bucknam, '81

Site Planning & N. A. Campagna, Jr. °67

Development F. W. Clark, '79

Soil/Rock Testing W. E. Hadge, '79

Geotechnlcal W. E. Joworakl, '73

Instrumentation C. A. Undberg, *78

Testing of Construction R. M. Simon, '72

Engineering for E. I. Steinberg, -so

Lateral Support T. vonRosenvlnge IV. '80

Systems

Rock Engineering The GEO Building

Groundwster 320 Noodham St.

Engineering Newton Upper

Underground Falls, MA 02164

Construction (617)467-8840

DebesCaip.

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Doslgn, Construction,

Management

Subsidiaries:

Charles N. Debes &

Assoc. Inc.

Alms Nelson Manor Inc.

Park Strathmoor

Corporation

Rockford Convalescent

Center Inc.

Chambro Corporation


S888 Strathmore Drlvo

Rockford, IL. 61107

EEHawkins&Sons

Company

Building contractor*

Steven H. HflwkJns< '57

186 Whiting Street

Hlrtgham. MA 02043

(617)749-6011

(617)749-6012

-

TECHNOLOGY REVIEW AH

A rigid arm pivots around the fixed

point A. At the end of the arm is a fol

lower (B) which runs in a curved track.

The track pivots about the fixed point

C. If AB = AC = r, find the shape of

the track such that its slope at C is al

ways vertical.

Speed Department

SD 1. Here's one from Smith D. Turner

(/dt):

Bill and Joe are to be paid $10 to wrap

and address a pile of packages. Joe ad

dresses one while Bill wraps one, but

Bill addresses three while Joe wraps one.

How should the $10 be divided between

them?

SD 2. We end with a bridge quickie from

Doug Van Patter:

South:

4k AK74

V KJ3

North:

AJ6

V AQ5

♦ A2 ♦ KJ54

+ AQ9 862 ♦ K 5

You are declarer (South) in a six-no-

trump contract. The opening lead of the

♦ 10 is taken with your ^J. Now you

wish you were in a grand slam. When

you cash the *K, West shows out. Can

you find any reasonable chance of still

making your contract?

Solutions

Y 1984. This is the same problem as Y 1985 (see

above) with only the one digit changed.

The following solution is from Harry Zaremba;

he points out that 36 numbers, shown with aster

isks, use the digits in the same order as the year

1984:

37 = (1 + 4) x 9 - 8

"38 = 19 x 8/4

39 = 48 - 9 x 1

40 = 41 - 9 + 8

•41 = 1 x 9 + 8 x 4

•42 = 1 + 9 + 8 X 4

43 = 91 - 48

•1 = 1*2 = 4/(18.-9)

•3-= 1* + 8/4•4 = 1-9 + 8 + 4

•5=1x9-8 + 4

•6=1+9-8 + 4

•7 = 19 - 8 - 4

•8=1 + 9-8/4

9 = 9xlH

10 = 9 - 1 + 8/4

•11 = 1 x 9 + 8/4

•12 = 1 + 9 + 8/4

13 = 94 - 81

•14 = 1 + 9 + 8 - 4

•15 = 19 - 8 + 4

16 = (9 - 1) x 8/4

•17 =19-8/4

•18 = 1 x 9 X 8/4

•19 = 1 + 9 x 8/4

•20 = (1 + 9) x 8/4

•21 = 19 + 8'4

•22 = 1+9 + 8 + 4

•23 = 19 + 8 - 4

24 = 41 - 9 - 8

25 =

26 =

27 =9x4-8-1

28 = 9x4-8x1

29 = 48 - 19

30 =

•31 = 19 + 8 + 4

32 = 81 - 49

•33 = 1° + 8 x 4

34 =

35 = 9 X (4 - I) + f

36 = 81/9 x 4

•44 = (19 - 8) x 4

45 = 81 - 9 x 4

46 =

47 = 48 - 1*48 = 89 - 41

49 - (1 + 4) x 8 -r 9

50 = 49 + I8•51 = 19 + 8 x 4

52 =

53 = (l+4)x9 + 8

54 = 18 + 9 x 4

55 =

56 = 48 - 1 + 9

57 = 98 - 41

58 = 41 + 9 + 8

59 = 91 - 8 x 4

60 = (9-l)x8-4

61 =

62 =

63 = 18 x 4 - 9

64 = (9 + 8-l)x4

65 = 84 - 19

66 =

67 = 48 + 19

•68 =1x9x8-4

•69 = 1 + 9 x 8 - 4

70 =

71 = 9 x 8 - 1*•72 = (1 + 9 + 8) x 4

73 = 9 x 8 + I4

74 = 84 - 9 - 1

75 = 89 - 14

•76 = 1 - 9 + 84

•77 =1+9x8 + 4

78 =

79 = 91 - 8 - 4

80 = (I4 + 9) x 8*81 = 1 x 91*11

•82 = 1 + 9<M>

83 = 84 - 1*

84 = 98 - 14

•85 = 1' + 8486 = 81 + 9 - 4

87 = 91 - 8 + 4

88 = 89 - I4

89 = 91 - 8/4

90 = 18 x (9 - 4)

91 = (9 + 4) x (8 - I)

92 = 89 + 4 - 1

•93 = 1 x 9 + 84

•94 = I + 9 + 84

•95 = 1 + 98 - 4

96 ■= (9 + 4 - 1) x 897 = 98 - I498 = 98 x I4

99 = (8 + 4 - 1) x 9100 =

Also solved by Avi Ornslein, George Aronson.

Harry (Hap) Hazard, Phelps Meaker, Joe Feil, Peter

Silverberg, Allan Tracht, and A. Holt.

A/S 1. Given the situation shown. While to move

and win.

Bert Daniels had a little trouble with this one:

1. N-B8, K-Rl (1 P-N3 loses to 2. P-B6, N-B2,

3. Q-K8 mate; while 1 K-Bl loses to 2. Q-K7

ch and Q-K8 mate);

2. Q-Q8 ch, N-Nl;

3. N-Q6 and the threat of smothered mate wins the

queen.

Also solved by Matthew Fountain, Kenneth Bern

stein, R. Hess, Avi Omstein, David Evans, David

Detlefs, George Aronson, Ronald Raines, Philip

Dangel, and the proposer, Robert Kimble.

A/S 2. An ordinary combination padlock requires

three ordered numbers to open, each between 0 and

39, inclusive. Thus there are 64,000 possible com

binations. If it is known that the sum of the three

numbers is 58 and the sum of the individual digits

of all three numbers is 13, how many combinations

are possible? If each of these possible combinations

is equally likely, what is the probability that the

second number is 34?

Many readers submitted computer programs that

calculated all possibilities. I preferred analyses that

reduced the number of possibilities to a manageable

level. Matthew Fountain actually submitted both a

program and an analysis. Here is the latter:

Let A equal the sum of the tens digits of the three

ordered numbers and B equal the sum of the units

3,2,0

3,1,1

2,2,1

Total

Permutations

6 (2)

3 (1)

_3 _

12 (3)

( ) indicates number of

permutations with 3 in

second position.

8,0.0

7.1,0

6,2,0

6,1,1

5,3,0

5,2,1

4,4,0

4.3,1

4.2.2

3,3,2

Permutations

3

6

6

3

6

6

3

6

3

3

(2)

(2)

(1)

Total 45 (5)

( ) indicates number of

permutations with 4 in

second position.

Nelson,Goulson&

Associates,Inc


Consultants


Servlcaa

Professional

Recruiting Services


All Disciplines

313 W. Hompden Ave.

Sulto 507

Englewood, CO 60110

(303) 761-7680

Other offices In

Albuquerque,

Colorado Spring),

Dallas & Seattle

Paul R. Coulson

PE/43

President

PaulEDutelle&

GompanyInc

Roofers snd Metal

Craftsmen

153 Pearl Street

Newton, MA 02158

LordElectricCompany

Inc.


to the nation since

Headquarters:

45 Rockefeller Plaza

Now York, N.V. 10111




Boston Office:

86 Coolidge Ave.

Watertown, MA 02172

(617) 926-5500

Alexander W. Mortal, Jr.


AlbanyInternationalResearchCo.

Contract RAD

for Industry in

Polymors

Fibers

Textiles

Plastics

Composites

Interpretive Testing

Chemical Analysis

Consulting

Expert Testimony

0. R. Petterson '59

E. R. Kaswoll '39

M. M. Platt '42

R. B. Davis '66

0. S. Brookateln '76M. A. Konney '83

1000 Providence

Highway

Dedham, MA 02026

617-326-5500

J. R. Dent '79

R. E. Erlandson '48

0. M. Boars 'B0

F. A. DITaranlo '83

LEA Gioup

UNENTHALEISENBERQ

ANDERSON, INC.ENGINEERS

Building Design

Environmental

Engineering

Sne/CMI DesignRoofing Technology

Consultants to

Industry, commerce,government and

Institutions.

75 Kneoland Street

Boston, MA 02111

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New York, NY

(212) 509-1922

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Anderson '50

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David A. Peters 77MSCE

Venfinesand

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Management

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solution ol complete,

difficult, new,

Infrequent, or uniqueproblems

Giorgio Plccagli,■67. Ph.D.

100 Dorado Terrace


(415)333-5084

Number

of sides

4

6

8

12

20

Edges

each

side

3

4

3

5

3

Total

edges

12

24

24

60

60

Number

of

dihedralsg

12

12

30

30

Dihedralsmeeting

at each

apex

4

4

3

5

NumbernfUl

apices

AV

20y

digits of the three ordered numbers. Then A t B= 13 and 10A + B » 58, with soluHons A = 5 andB =■ 8. The table at the bottom of page A25 showsthat A can be decomposed into three ordered digits,none exceeding 3, in twelve ways, and B can be

decomposed into three ordered digits 45 ways. Thetotal combinations are 12 x 45 = 540. Those with34 as second number are 3 x 5 = 15. The probabilitythat 34 is the second number is 15/540 = 1/36.Also solved by Kenneth Bernstein, Richard Hess

Avi Omstein, David Evans, David Detlefs, GeorgeAronson, Dennis Sandow, Rita Carp, Gerry Cross-man, Harry Zaremba, Richard Marks, WinslowHartford, Yale Zussman, P. Jung, Frank Carbin,Steve Feldman, Aaron Hirschberg, Dave Mohr,Alfred Anderson, Thomas Stowe, and the proposerJohn Prussing.

A/S 3. Fill in the missing entry in the table abovepertaining to regular polyhedra. Can the values inthe last column be determined by a formula?

The table contained two typos: a square has eightapices and three dihedrals meeting at each apex.These errors did not seem to cause much trouble.In particular, Avi Omstein submitted the following:The missing number is 12 apices for the icosa-

hedron. The number of apices is given by the following:

(number of sides)(edges on each side)/(dihedralsmeeting at each apex)or

(total edges)/(dihedrals meeting a I each apex)or

2(number of dihcdrals)/(dihedrals meeting at eachapex).

Also solved by Matthew Fountain, Kenneth Bem-slein, Richard Hess, David Evans, David DetlefsGerry Grossman, Harry (Hap) Hazard, Harry Zaremba, Richard Marks, Winslow Hartford YaleZussman, and Albert Mullin.

A/S 4. Find infinitely many positive integers n notcontaining the digit zero such that nJ - 1 containsjust two digits neither of which is zero The dicitsmay be repeated.

Jerry Marks sent us three patterns:7

67

667

5667

66667

5

65

665

6665

48

4488

444888

44448888

4444488888

24

4224

442224

44422224

34

334

3334

33334

15

1155

111555

11113555

1111155555

66665 444422224

,-> RiC'j™tHcSS "a,S a Proof of one of these patterns:(2 x 10-/3 + 1/3)' - 1 = 4 x 10*V9 + 4 x 10"/9+1/9-1

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= 4(102n - l)/9 + 4(10" - l)/9 + 4/9 + 4/9 - 8/9= 2n 4's + n 4's

= n 4's followed by n 8's.

Also solved by Kenneth Bernstein, David Evans,David Detlefs, Dennis Sandow, Gerry Grossman,Harry Zaremba, and the proposer, Matthew Fountain.

A/S 5. Given a triangle ABC, draw its incircle andconsider triangle DEF determined by the points oftangency. Show that the area of triangle DEF is (r/

d)A, where r is the radius of the incircle, d is thediameter of the dreumcirele, and A is the area oftriangle ABC.

We give two different solutions, the first fromKenneth Bernstein and the second from PhelpsMeaker:

Bernstein begins by letting the sides of triangle ABCbe a, b, jind c. Define k by:k ■= V{a + b + cx-a + b + cXa - b + c«a + b - c)

The radius, R, of the circumcircle is abc/k. The area

of triangle ABC is k/4. The radius, r, of the incirdeis k/2(a + b + c). Denote the side FE of triangleFED by a'. Then a' can be expressed in terms of rand angle A:

(a')J = 2^(1 + (cos A)).

The term (cos A) can be expressed in terms of a, b,and c using the law of cosines:a2 = b2 + c3 - 2bc cos A.Combining the last two expressions:a' = (-a + btc)8x V (a - b + c)(a + b -a (a + btc)8x V (a - b + c)(a + b - cybc.with similar expressions for b' and c\ The area oftriangle FED is k'/4 where k' is defined similarly tok with a' substituted for a, etc. After much algebra,the area of triangle FED is

k(-a + b + c)(a - b + c)(a + b - c)/16abc= k'/16abc(a + b + c)= (r/2R)(area ABC).

To establish his solution, Phelps Meaker lets Hequal the altitude AD of triangle ABC; O is the water of the incircle; G is the cenler of the circumcircle-and a is one-half of the apical angle. Then, withrespect to triangle AEO

r = EO = OD = (H - r) sin a ;r + r sin a = H sin a; andr = H sin a/(l + sin a).With respect to triangle ABD-AJ = AB/2 = H/2 cos a;

JANUARY 1985

AG = AJ/COS a = H/2 cos2 a;

d = 2AC = H/cos2 a.

The area of triangle ABC is

H X H Ian a = H' sin a/cos a.

With respect to triangle EDF:

EP = EO cos a; angle PEO = a; OP = EO sin a.

Then the area of triangle EDF is given by

EP x (OD + OP) = r cos a (r + r sin a)

= r2 cos a + r1 sin a cos a = r^ cos a (1 + sin a).

Substituting for r, the area of triangle EDF is given

by

H2 sin2 a cos a (1 + sin a)/(l + sin a)!= H2 sin2 a cos a/(l + sin a).

(r/d)A = (H sin a/(l + sin a)l[cos2 a/H][H2 sin a/

cos a]

= H2 sin2 a cos a/(l + sin a).Also solved by Matthew Fountain, Richard Hess,

David Evans, Robert Hollenbach, Henry Lieber-

man, Mary Linderman, Howard Stern, and the pro

poser, Harry Zaremba.


1983 JUL 5. Matthew Fountain sent us the follow

ing:

Donald Savage's comments on 1983 JUL 5 stimu

lated me to further investigation. I found (a) there

are 29(2)" n-digit numbers whose squares are suit

able with respect to their last n digits and (b) there

are 29(2)n(0.2)' n-digit numbers whose squares are

suitable with respect to their last n + r digits; (a) is

exact when n>3; (b) is an excellent approximation

when n is large and r small. Both (a) and (b) are

consistent with the notion that the middle digits of

squares are representative of random numbers, ex

cept that (a) is more than chance. I wrote a Pascal

program that generated all the (a) numbers up to

and including those of 20 digits and printed out

those that came closest to having entirely suitable

squares. My IBM took about 2.5 days to cover the

range of 15 through 20 digits. 1 found that the results

differed from (b) because of the pecularity of two

sequences of digits. For example, (b) expects there

to be 3.1 numbers of 20 digits or less with squares

having their last 30 digits suitable. Actually there

are 22. But 15 differ from 83,333,333,333,333,333,332

only in the first six or less digits, and four differ

from 21,666,666,666,666,666,662 in the first five or

less digits. Only 78,537,356,970,849,674,736 appears

to resemble a random number. My conclusion is

that (b) is probably a good estimate of the odds that

there exists a large n-digit number with an (n + r)-

digit square, all of whose digits are suitable. The

pecularity of the two sequences should not affect

(b) when r = n or r = (n - 1). The odds, of course,

seem very small. It is interesting that Los Alamos

in its early days took sequences of digits from the

middles of consecutive squares as random num

bers, a suggestion of von Neumann. Some cyclic

patterns were observed, the worst being too many

consecutive zeroes.

19S4 APR 1. Harry (Hap) Hazard points out that

spades can be led or steel but not lead.

APR 2. Richard Halloran has responded.

M/J.5. Dayton Datlowe has responded.

JUL 2. Ivor Morgan and Mary Lindenberg have re

sponded.

A/S SD2. Arthur Carp, Ronald Martin, and Gordon

Thomas report that pan's dimensions are 7 x 5 x

1.5 inches.


SD 1. The editor needs to assume that the speed

ratio between Joe and Bill is constant for wrapping

and addressing. In that case, the ratio is V3 and

Joe should receive

$10/(1 + V3).

SD 2. Lead low toward the A], and hope West has

the *Q. (My partner found this line of play, but

most declarers went down.)

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Giorgio Plccagll,

Ph.D., '67

100 Dorado Terrace

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Massachusetts Institute

of Technology

Report of the PresidentFor the Academic Year

1983-1984

The idea of MIT began in the mind of one man —

William Barton Rogers— and took root in a fertile setting:

in nineteenth-century America, in Boston. It was an idea

that captured the imagination and intellectual energy of

those citizens who saw the need for a new kind of ed

ucation which would emphasize, in Rogers's words,

"... the value of science in its great modern applications

to the practical arts of life, to human comfort, and health,

and to social wealth and power." The founding idea

was accompanied by an impulse, a spirit, which was

just as revolutionary as the idea itself: the spirit of in

venting the future. Rogers was not bound by the tra

ditional ways of organizing and seeking knowledge; his

entrepreneurial spirit and willingness to take risks in

pursuit of an exciting idea are the MIT tradition.

As MIT has grown, it has embraced and invented a

host of intellectual domains. Our departments and

academic programs reflect in their names and in their

activities the remarkable ability of the faculty to anticipate

and shape the future. Scores of interdepartmental re

search laboratories and centers constitute perhaps an

even more sensitive barometer of the ways in which

MIT transcends the boundaries of tradition in both or

ganization and style.

Today, our fields of study and scholarship include

management, urban planning, humanities and the social

sciences, as well as the natural sciences, architecture,

and engineering. MIT has become more than an institute

of technology in the nineteenth-century sense of that

phrase. To use the words of James R. Killian, Jr., first

spoken in 1949, MIT is "a university polarized around

science, engineering, and the arts." And yet the founding

idea of MIT continues and is central to our future, for

we are now and must remain the strongest science-based

research university in the world. Our historic commit

ment to scientific and quantitative methods remains at

the core of our approach to learning; it permeates the

whole spectrum of our degree programs, and is a touch

stone of common interest and purpose for all who study,

teach, and work here.

Given this foundation, MIT's current mission can be

stated succinctly: to provide the highest quality programs

of education and research in each of those areas of study

and investigation in which we have developed strength

and competence, and to do so with a strong commitment

to public service and to a diversity of backgrounds, in

terests, and points of view among the faculty, students,

and staff.

A28 JANUARY 1985 PHOTOGRAPH OF WILLIAM BARTON ROGERS: M.IT MUSEUM

MidsummerSnowstorm - New York Universitygottlieb/tr/back-issues/1980s/1985/1985...MidsummerSnowstorm Ourgroup is just nowmovingto anewlyrenovatedfloor. Wehad a major influence in specifying

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